The Material Balance for Chemical Reactors
Copyright c
2015 by Nob Hill Publishing, LLC
1
The Material Balance for Chemical Reactors
Copyright c
2015 by Nob Hill Publishing, LLC
1
General Mole Balance
R
j
V
Q
1
c
j
1
Q
0
c
j
0
Conservation of mass







rate of
accumulation
of component
j







=
(
rate of inflow
of component
j
)
−
(
rate of outflow
of component
j
)
+







rate of generation
of component
j
by
chemical reactions







(1)
2
General Mole Balance
R
j
V
Q
1
c
j
1
Q
0
c
j
0
Conservation of mass
d
dt
Z
V
c
j
dV
=
Q
0
c
j
0
−
Q
1
c
j
1
+
Z
V
R
j
dV
(2)
3
General Mole Balance
d
dt
Z
V
c
j
dV
=
Q
0
c
j
0
−
Q
1
c
j
1
+
Z
V
R
j
dV
•
Equation 2 applies to every chemical component in the system,
j
=
1
,
2
, . . . , n
s
, including inerts, which do not take place in any reactions.
•
Assuming component
j
enters and leaves the volume element only by con-
vection with the inflow and outflow streams, i.e. neglecting diffusional flux
through the boundary of the volume element due to a concentration gradient.
•
The diffusional flux will be considered during the development of the material
balance for the packed-bed reactor.
4
Rate expressions
•
To solve the reactor material balance, we require an expression for the pro-
duction rates,
R
j
R
j
=
X
i
ν
ij
r
i
•
Therefore we require
r
i
as a function of
c
j
•
This is the subject of chemical kinetics, Chapter 5
•
Here we use common reaction-rate expressions without derivation
5
The Batch Reactor
R
j
•
The batch reactor is assumed well stirred
6
•
Let the entire reactor contents be the reactor volume element
7
Batch Reactor
d
dt
Z
V
c
j
dV
=
Q
0
c
j
0
−
Q
1
c
j
1
+
Z
V
R
j
dV
•
Because the reactor is well stirred, the integrals in Equation 2 are simple to
evaluate,
Z
V
R
c
j
dV
=
c
j
V
R
Z
V
R
R
j
dV
=
R
j
V
R
•
The inflow and outflow stream flowrates are zero,
Q
0
=
Q
1
=
0
.
d
c
j
V
R
dt
=
R
j
V
R
(3)
8
Reactor Volume
•
Equation 3 applies whether the reactor volume is constant or changes during
the reaction.
•
If the reactor volume is constant (liquid-phase reactions)
dc
j
dt
=
R
j
(4)
•
Use Equation 3 rather than Equation 4 if the reactor volume changes signifi-
cantly during the course of the reaction.
9
Analytical Solutions for Simple Rate Laws
•
In general the material balance must be solved numerically.
•
If the reactor is isothermal, we have few components, the rate expressions
are simple, then analytical solutions of the material balance are possible.
•
We next examine derive analytical solutions for some classic cases.
10
First-order, irreversible
•
Consider the first-order, irreversible reaction
A
k
-
→
B
,
r
=
kc
A
(5)
•
The material balance for a constant-volume reactor gives
dc
A
dt
= −
kc
A
(6)
•
Watch the sign!
11
First-order, irreversible
•
We denote the initial concentration of A as
c
A
0
,
c
A
(t)
=
c
A
0
,
t
=
0
•
The solution to the differential equation with this boundary condition is
c
A
=
c
A
0
e
−
kt
(7)
12
First-order, irreversible
0
0.2
0.4
0.6
0.8
1
0
1
2
3
4
5
k
=
0
.
5
k
=
1
k
=
2
k
=
5
cA
cA
0
t
13
First-order, irreversible
•
The A concentration decreases exponentially from its initial value to zero with
increasing time.
•
The rate constant determines the shape of this exponential decrease. Rear-
ranging Equation 7 gives
ln
(c
A
/c
A
0
)
= −
kt
14
First-order, irreversible
0.001
0.01
0.1
1
0
1
2
3
4
5
k
=
0
.
5
k
=
1
k
=
2
k
=
5
cA
cA
0
t
•
One can get an approximate value of the rate constant from the slope of the
straight line.
•
This procedure is a poor way to determine a rate constant and should be
viewed only as a rough approximation (Chapter 9).
15
First-order, irreversible
The B concentration is determined from the A concentration.
1. Solve the material balance for component B,
dc
B
dt
=
R
B
=
kc
A
(8)
with the initial condition for B,
c
B
(
0
)
=
c
B
0
2. Note that the sum of
c
A
and
c
B
is constant.
d(c
A
+
c
B
)
dt
=
R
A
+
R
B
=
0
Therefore,
c
A
+
c
B
is a constant.
16
First-order, reversible
•
The value is known at
t
=
0
,
c
A
+
c
B
=
c
A
0
+
c
B
0
•
So we have an expression for
c
B
c
B
=
c
A
0
+
c
B
0
−
c
A
(9)
c
B
=
c
B
0
+
c
A
0
(
1
−
e
−
kt
)
(10)
17
First-order, reversible
•
Consider now the same first-order reaction, but assume it is reversible
A
k
1
-
*
)
-
k
−
1
B
(11)
•
The reaction rate is
r
=
k
1
c
A
−
k
−
1
c
B
.
•
The material balances for A and B are now
dc
A
dt
= −
r
= −
k
1
c
A
+
k
−
1
c
B
c
A
(
0
)
=
cA
0
dc
B
dt
=
r
=
k
1
c
A
−
k
−
1
c
B
c
B
(
0
)
=
c
B
0
•
Notice that
c
A
+
c
B
=
c
A
0
+
c
B
0
remains constante
18
First-order, reversible
•
Eliminate
c
B
in the material balance for A gives
dc
A
dt
= −
k
1
c
A
+
k
−
1
(c
A
0
+
c
B
0
−
c
A
)
(12)
•
How do we want to solve this one?
•
Particular solution and homogeneous solution (see text)
•
Laplace transforms (control course)
•
Separation!
19
First-order, reversible
dc
A
dt
=
ac
A
+
b
Z
c
A
c
A
0
dc
A
ac
A
+
b
=
Z
t
0
dt
1
a
ln
(ac
A
+
b)
c
A
c
A
0
=
t
c
A
=
c
A
0
e
at
−
b
a
(
1
−
e
at
)
Substitute
a
= −
(k
1
+
k
−
1
)
,
b
=
k
−
1
(c
A
0
+
c
B
0
)
20
First-order, reversible
c
A
=
c
A
0
e
−
(k
1
+
k
−
1
)t
+
k
−
1
k
1
+
k
−
1
(c
A
0
+
c
B
0
)
h
1
−
e
−
(k
1
+
k
−
1
)t
i
(13)
•
The B concentration can be determined by switching the roles of A and B and
k
1
and
k
−
1
in Reaction 11, yielding
c
B
=
c
B
0
e
−
(k
1
+
k
−
1
)t
+
k
1
k
1
+
k
−
1
(c
A
0
+
c
B
0
)
h
1
−
e
−
(k
1
+
k
−
1
)t
i
(14)
21
First-order, reversible
0
0.2
0.4
0.6
0.8
1
0
1
2
3
4
5
cB(t)
cA(t)
cAs
cBs
c
t
Figure 1: First-order, reversible kinetics in a batch reactor,
k
1
=
1
,
k
−
1
=
0
.
5
,
c
A
0
=
1
,
c
B
0
=
0
.
22
Nonzero steady state
•
For the reversible reaction, the concentration of A does not go to zero.
•
Taking the limit
t
-
→ ∞
in Equation 13 gives
c
As
=
k
−
1
k
1
+
k
−
1
(c
A
0
+
c
B
0
)
in which
c
As
is the steady-state concentration of A.
23
Nonzero steady state
•
Defining
K
1
=
k
1
/k
−
1
allows us to rewrite this as
c
As
=
1
1
+
K
1
(c
A
0
+
c
B
0
)
•
Performing the same calculation for
c
B
gives
c
Bs
=
K
1
1
+
K
1
(c
A
0
+
c
B
0
)
24
Second-order, irreversible
•
Consider the irreversible reaction
A
k
-
→
B
(15)
in which the rate expression is second order,
r
=
kc
2
A
.
•
The material balance and initial condition are
dc
A
dt
= −
kc
2
A
,
c
A
(
0
)
=
c
A
0
(16)
•
Our first nonlinear differential equation.
25
Second-order, irreversible
•
Separation works here
dc
A
c
2
A
= −
kdt
Z
c
A
c
A
0
dc
A
c
2
A
= −
k
Z
t
0
dt
1
c
A
0
−
1
c
A
= −
kt
•
Solving for
c
A
gives
c
A
=
1
c
A
0
+
kt
−
1
(17)
•
Check that this solution satisfies the differential equation and initial condition
26
Second-order, irreversible
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
1
2
3
4
5
first order
second order
cA
cA
0
•
The second-order reaction decays
more slowly
to zero than the first-order
reaction.
27
Another second-order, irreversible
A
+
B
k
-
→
C
r
=
kc
A
c
B
(18)
•
The material balance for components A and B are
dc
A
dt
= −
r
= −
kc
A
c
B
(19)
dc
B
dt
= −
r
= −
kc
A
c
B
(20)
•
Subtract B’s material balance from A’s to obtain
d(c
A
−
c
B
)
dt
=
0
28
Another second-order, irreversible
•
Therefore,
c
A
−
c
B
is constant, and
c
B
=
c
A
−
c
A
0
+
c
B
0
(21)
•
Substituting this expression into the material balance for A yields
dc
A
dt
= −
kc
A
(c
A
−
c
A
0
+
c
B
0
)
•
This equation also is separable and can be integrated to give (you should
work through these steps),
c
A
=
(c
A
0
−
c
B
0
)
1
−
c
B
0
c
A
0
e
(c
B
0
−
c
A
0
)kt
−
1
,
c
A
0
â‰
c
B
0
(22)
29
Another second-order, irreversible
•
Component B can be computed from Equation 21, or by switching the roles
of A and B in Reaction 18, giving
c
B
=
(c
B
0
−
c
A
0
)
1
−
c
A
0
c
B
0
e
(c
A
0
−
c
B
0
)kt
−
1
(23)
•
What about component C? C’s material balance is
dc
C
dt
=
kc
A
c
B
and therefore,
d(c
A
+
c
C
)/dt
=
0
. The concentration of C is given by
c
C
=
c
A
0
−
c
A
+
c
C
0
30
Another second-order, irreversible
•
Notice that if
c
A
0
> c
B
0
(Excess A), the steady state
c
As
=
c
A
0
−
c
B
0
c
Bs
=
0
c
Cs
=
c
B
0
+
c
C
0
•
For
c
B
0
> c
A
0
(Excess B), the steady state is
c
As
=
0
c
Bs
=
c
B
0
−
c
A
0
c
Cs
=
c
A
0
+
c
C
0
31
n
th-order, irreversible
The
n
th-order rate expression
r
=
kc
n
A
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
0
0.5
1
1.5
2
2.5
3
r
n
=
3
2
1
1
/
2
0
cA
32
n
th-order, irreversible
A
k
-
→
B
r
=
kc
n
A
dc
A
dt
= −
r
= −
kc
n
A
•
This equation also is separable and can be rearranged to
dc
A
c
n
A
= −
kdt
•
Performing the integration and solving for
c
A
gives
c
A
=
h
c
−
n
+
1
A
0
+
(n
−
1
)kt
i
1
−
n
+
1
,
n
â‰
1
33
n
th-order, irreversible
•
We can divide both sides by
c
A
0
to obtain
c
A
c
A
0
=
[
1
+
(n
−
1
)k
0
t]
1
−
n
+
1
,
n
â‰
1
(24)
in which
k
0
=
kc
n
−
1
A
0
has units of inverse time.
34
n
th-order, irreversible
0
0.2
0.4
0.6
0.8
1
0
1
2
3
4
5
1
2
3
4
n
=
5
cA
cA
0
t
•
The larger the value of
n
, the more slowly the A concentration approaches
zero at large time.
35
n
th-order, irreversible
•
Exercise care for
n <
1
,
c
A
reaches zero in
finite time
.
0
0.2
0.4
0.6
0.8
1
0
0.5
1
1.5
2
cA
cA
0
n
=
2
1
1
/
2
0
−
1
/
2
−
1
−
2
t
36
Negative order, inhibition
•
For
n <
0
, the rate decreases with increasing reactant concentration; the
reactant
inhibits the reaction
.
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
0
0.5
1
1.5
2
2.5
3
r
0
−
1
/
2
−
1
n
=−
2
cA
37
Negative order, inhibition
•
Inhibition reactions are not uncommon, but watch out for small concentra-
tions. Notice the rate becomes unbounded as
c
A
approaches zero, which is
not physically realistic.
•
When using an ODE solver we may modify the right-hand sides of the material
balance
dc
A
dt
=
(
−
kc
n
A
,
c
A
>
0
0
,
c
A
=
0
•
Examine the solution carefully if the concentration reaches zero.
38
Two reactions in series
•
Consider the following two irreversible reactions,
A
k
1
-
→
B
(25)
B
k
2
-
→
C
(26)
•
Reactant A decomposes to form an intermediate B that can further react to
form a final product C.
•
Let the reaction rates be given by simple first-order rate expressions in the
corresponding reactants,
r
1
=
k
1
c
A
r
2
=
k
2
c
B
39
Two reactions in series
•
The material balances for the three components are
dc
A
dt
=
R
A
= −
r
1
= −
k
1
c
A
dc
B
dt
=
R
B
=
r
1
−
r
2
=
k
1
c
A
−
k
2
c
B
dc
C
dt
=
R
C
=
r
2
=
k
2
c
B
•
The material balance for component A can be solved immediately to give
c
A
=
c
A
0
e
−
k
1
t
as before.
40
Two reactions in series – B
•
The material balance for B becomes
dc
B
dt
+
k
2
c
B
=
k
1
c
A
0
e
−
k
1
t
Oops, not separable, now what?
•
Either Laplace transform or particular solution, homogeneous equation ap-
proach produces
c
B
=
c
B
0
e
−
k
2
t
+
c
A
0
k
1
k
2
−
k
1
h
e
−
k
1
t
−
e
−
k
2
t
i
,
k
1
â‰
k
2
(27)
41
Two reactions in series – C
•
To determine the C concentration, notice from the material balances that
d(c
A
+
c
B
+
c
C
)/dt
=
0
. Therefore,
c
C
is
c
C
=
c
A
0
+
c
B
0
+
c
C
0
−
c
A
−
c
B
42
Two reactions in series
0
0.2
0.4
0.6
0.8
1
0
1
2
3
4
5
cA(t)
cB(t)
cC(t)
c
t
Figure 2: Two first-order reactions in series in a batch reactor,
c
A
0
=
1
,
c
B
0
=
c
C
0
=
0
,
k
1
=
2
,
k
2
=
1
.
43
Two reactions in parallel
•
Consider next two parallel reactions of A to two different products, B and C,
A
k
1
-
→
B
(28)
A
k
2
-
→
C
(29)
•
Assume the rates of the two irreversible reactions are given by
r
1
=
k
1
c
A
and
r
2
=
k
2
c
A
.
44
Two reactions in parallel
•
The material balances for the components are
dc
A
dt
=
R
A
= −
r
1
−
r
2
= −
k
1
c
A
−
k
2
c
A
dc
B
dt
=
R
B
=
r
1
=
k
1
c
A
dc
C
dt
=
R
C
=
r
2
=
k
2
c
A
45
Two reactions in parallel
•
The material balance for A can be solved directly to give
c
A
=
c
A
0
e
−
(k
1
+
k
2
)t
(30)
•
Substituting
c
A
(t)
into B’s material balance gives
dc
B
dt
=
k
1
c
A
0
e
−
(k
1
+
k
2
)t
•
This equation is now separable and can be integrated directly to give
c
B
=
c
B
0
+
c
A
0
k
1
k
1
+
k
2
1
−
e
−
(k
1
+
k
2
)t
(31)
46
Two reactions in parallel
•
Finally, component C can be determined from the condition that
c
A
+
c
B
+
c
C
is constant or by switching the roles of B and C, and
k
1
and
k
2
in Equation 31,
c
C
=
c
C
0
+
c
A
0
k
2
k
1
+
k
2
1
−
e
−
(k
1
+
k
2
)t
(32)
47
Two reactions in parallel
0
0.2
0.4
0.6
0.8
1
0
0.5
1
1.5
2
2.5
3
cA(t)
cB(t)
cC(t)
c
t
Figure 3: Two first-order reactions in parallel in a batch reactor,
c
A
0
=
1
,
c
B
0
=
c
C
0
=
0
,
k
1
=
1
,
k
2
=
2
.
48
Two reactions in parallel
•
Notice that the two parallel reactions
compete
for the same reactant, A
•
The rate constants determine which product is favored
•
Large values of
k
1
/k
2
favor the formation of component B compared to C and
vice versa
49
Conversion, Yield, Selectivity
There are several ways to define selectivity, yield and conversion, so be clear
about the definition you choose.
Point selectivity:
The point (or instantaneous) selectivity is the ratio of the production rate of
one component to the production rate of another component.
Overall selectivity:
The overall selectivity is the ratio of the amount of one component pro-
duced to the amount of another component produced.
Yield:
The yield of component
j
is the fraction of a reactant that is converted into component
j
.
Conversion:
Conversion is normally defined to be the fraction of a component that has been
converted to products by the reaction network. Conversion has several definitions and
conventions. It is best to state the definition in the context of the problem being solved.
50
The Continuous-Stirred-Tank Reactor (CSTR)
Q
c
j
Q
f
c
jf
R
j
51
Writing the material balance for this reactor gives
d
c
j
V
R
dt
=
Q
f
c
jf
−
Qc
j
+
R
j
V
R
,
j
=
1
, . . . , n
s
(33)
52
CSTR – Constant Density
•
If the reactor volume is constant and the volumetric flowrates of the inflow
and outflow streams are the same, Equation 33 reduces to
dc
j
dt
=
1
Ï„
(c
jf
−
c
j
)
+
R
j
(34)
•
The parameter
Ï„
=
V
R
/Q
f
is called the
mean residence time
of the CSTR.
•
We refer to this balance as the constant-density case. It is often a good ap-
proximation for liquid-phase reactions.
53
CSTR – Steady State
•
The steady state of the CSTR is described by setting the time derivative in
Equation 33 to zero,
0
=
Q
f
c
jf
−
Qc
j
+
R
j
V
R
(35)
•
Conversion of reactant
j
is defined for a steady-state CSTR as follows
x
j
=
Q
f
c
jf
−
Qc
j
Q
f
c
jf
(steady state)
(36)
•
One can divide Equation 35 through by
V
R
to obtain for the constant-density
case
c
j
=
c
jf
+
R
j
Ï„
(steady state, constant density)
(37)
54
Transient behavior of the CSTR
•
Consider a first-order, liquid-phase reaction in an isothermal CSTR
A
k
-
→
2B
r
=
kc
A
the feed concentration of A is
c
Af
=
2
mol/L, the residence time of the reactor
is
Ï„
=
100
min, and the rate constant is
k
=
0
.
1
min
−
1
.
1. Find the steady-state concentration of A in the effluent for the given feed.
2. Plot the concentration of A versus time for constant feed concentration
c
Af
=
2
mol/L if
the reactor is initially filled with an inert so
c
A
0
=
0
mol/L.
3. Plot the concentration of A versus time for constant feed concentration
c
Af
=
2
mol/L if
the reactor is initially filled with feed so
c
A
0
=
2
mol/L.
55
Transient CSTR Solution. Part 1
•
Liquid phase: assume the fluid density is constant.
c
A
=
c
Af
+
R
A
Ï„
•
Substituting the production rate
R
A
= −
kc
A
and solving for
c
A
gives the
steady-state concentration
c
As
=
c
Af
1
+
kτ
•
Substituting in the numerical values gives
c
As
=
2
mol/L
1
+
(
0
.
1
min
−
1
)(
100
min
)
=
0
.
182
mol/L
56
Parts 2 and 3
dc
A
dt
=
1
Ï„
c
Af
−
c
A
−
kc
A
(38)
c
A
(
0
)
=
c
A
0
•
This equation is also separable. The analytical solution is
c
A
(t)
=
c
A
0
e
−
(
1
/Ï„
+
k)t
+
c
Af
1
+
kτ
h
1
−
e
−
(
1
/Ï„
+
k)t
i
(39)
57
Parts 2 and 3
0
0.5
1
1.5
2
0
10
20
30
40
50
60
70
80
c
A
(t
)
(mol/L)
cA
0
=
0
cA
0
=
2
cAs
=
0
.
182
t
(min)
•
Both solutions converge to the same steady-state even though the starting
conditions are quite different.
58
Phenol production in a CSTR
•
Consider the reaction of cumene hydroperoxide (CHP) to phenol and acetone
(
C
6
H
5
)
C
(
CH
3
)
2
OOH
-
→
(
C
6
H
5
)
OH
+
(
CH
3
)
2
CO
(40)
CHP
-
→
phenol
+
acetone
(41)
•
The reaction is pseudo-first-order
r
=
kc
CHP
•
Find the reactor volume to achieve 85% conversion of CHP at steady state.
The flowrate into the reactor is
Q
f
=
26
.
9 m
3
/
hr
and
k
=
4
.
12 hr
−
1
.
59
Phenol production
•
Liquids at 85
â—¦
C
, so assume constant density and
Q
=
Q
f
.
c
A
=
c
Af
+
R
A
Ï„
R
A
= −
kc
A
, and solving for the CHP concentration gives
c
A
=
c
Af
1
+
kτ
(42)
60
Phenol production
•
The conversion of CHP (for
Q
=
Q
f
) is
x
A
=
c
Af
−
c
A
c
Af
=
1
−
c
A
c
Af
x
A
=
kτ
1
+
kτ
•
Solving this equation for
Ï„
gives
Ï„
=
1
k
x
A
1
−
x
A
61
Phenol production
•
Substituting the relation
Ï„
=
V
R
/Q
f
and solving for
V
R
gives
V
R
=
Q
f
x
A
k(
1
−
x
A
)
•
Substituting in the known values gives the required CSTR volume
V
R
=
(
26
.
9 m
3
/
hr
)(
0
.
85
)
(
4
.
12 hr
−
1
)(
0
.
15
)
=
37 m
3
62
The Semi-Batch Reactor
•
The semi-batch reactor is a cross between the batch reactor and CSTR.
•
The semi-batch reactor is initially charged with reactant, like the batch reac-
tor, but allows a feed addition policy while the reaction takes place, like the
CSTR.
•
Normally there is no outflow stream.
63
The Semi-Batch Reactor
•
We set
Q
=
0
in the CSTR material balance to obtain
d
c
j
V
R
dt
=
Q
f
c
jf
+
R
j
V
R
,
j
=
1
, . . . , n
s
(43)
•
One may choose to operate a semi-batch reactor to control the reaction rate
or heat release during reaction by slowly adding one of the reactants in the
feed stream.
•
Compared to the batch reactor, the semi-batch reactor provides more com-
plete use of the reactor volume in reactions such as polymerizations that
convert from lower density to higher density during the course of the reac-
tion.
64
Volume Change Upon Reaction
d
c
j
V
R
dt
=
Q
f
c
jf
−
Qc
j
+
R
j
V
R
(44)
•
Equation 44 covers both the batch, CSTR and semi-batch reactors, depending
on how we specify
Q
f
and
Q
.
•
If we multiply Equation 44 by the molecular weight of species
j
and sum over
all species we obtain,
d(
P
j
c
j
M
j
V
R
)
dt
=
Q
f
X
j
c
jf
M
j
−
Q
X
j
c
j
M
j
+
X
j
R
j
M
j
V
R
(45)
•
The term
P
j
c
j
M
j
is simply the mass density of the reactor contents, which
65
we denote
Ï
Ï
=
n
s
X
j
=
1
c
j
M
j
(46)
66
Volume Change Upon Reaction
•
The term
P
j
c
jf
M
j
is the mass density of the feedstream,
Ï
f
.
•
We know that conservation of mass in chemical reactions implies
P
j
R
j
M
j
=
0
(see Chapter 2). Substitution into Equation 45 leads to
d(ÏV
R
)
dt
=
Q
f
Ï
f
−
QÏ
(47)
•
Equation 47 is clearly a total mass balance, in which the total mass in the
reactor changes in time due to the inflow and outflow of mass.
•
Notice that chemical reactions play no role in the total mass balance.
67
Equation of state for the mixture
•
If we have a single-phase system at equilibrium, the intensive variables
c
j
,
T
,
P
, completely specify all intensive variables of the system.
•
In this chapter we consider
T
and
P
to be known, fixed quantities. Therefore,
the density of the reaction mixture, which is an intensive variable, is known
if the
c
j
are known.
•
This relationship is one form of the equation of state for the mixture
Ï
=
e
f (T , P , c
1
, c
2
, . . . , c
n
s
)
•
Substituting the definition of density, we can express the equation of state
as
f (c
1
, c
2
, . . . , c
n
s
)
=
0
68
Equation of state for the mixture
•
For example, we could express the equation of state in terms of the partial
molar volumes as
X
j
c
j
V
j
=
1
in which
V
j
is the partial molar volume of component
j
in the mixture.
•
The partial molar volumes are functions of
T
,
P
and
c
j
.
69
Equation of state for the mixture — Ideal mixture
•
If we assume an ideal mixture, this reduces to
X
j
c
j
V
â—¦
j
=
1
,
ideal mixture
in which
V
â—¦
j
is the specific volume of pure component
j
, which is a function
of only
T
and
P
.
•
We assume that a thermodynamic equation of state is valid even when the
reactor is not at equilibrium.
70
Constant density
•
Because the reaction mixture density,
Ï
, is independent of composition, it
does not vary with time either and we can set it to the feed value,
Ï
=
Ï
f
(48)
•
The total mass balance then reduces to
dV
R
dt
=
Q
f
−
Q
(49)
which is sometimes referred to as a “volume balance.â€
•
This terminology should be avoided.
71
Constant density
dV
R
dt
=
Q
f
−
Q
•
Batch reactor. For the batch reactor,
Q
=
Q
f
=
0
. We can therefore conclude
from Equation 49 that a batch reactor with constant density has constant
volume.
•
CSTR (dynamic and steady state). If the outflow of the CSTR is regulated so
that the CSTR has constant volume, then we can conclude from Equation 49
that
Q
=
Q
f
.
•
Semi-batch reactor. In the semi-batch reactor, the reactor is filled during
operation so
Q
f
is specified and positive for some time and
Q
=
0
. The
solution to Equation 49 then determines the change in volume of the reactor
during the filling operation.
72
Nonconstant density
Unknowns.
•
In the general case, consider the following variables to fully determine the
state of the reactor:
T , P , n
j
, V
R
.
•
We also require the value of
Q
to specify the right-hand sides of the material
balances.
•
The set of unknowns is
n
j
, V
R
, Q
.
•
We therefore have
n
s
+
2
unknowns.
Equations.
•
We have the
n
s
equations from the component mole balances.
•
The equation of state provides one additional equation.
•
The final equation is provided by a statement of reactor operation.
73
Nonconstant density – reactor operation
1. Constant-volume reactor. The constant-volume reactor can be achieved by
allowing overflow of the reactor to determine flowrate out of the reactor. In
this situation,
V
R
is specified as the additional equation.
2. Constant-mass reactor. The constant-mass reactor can be achieved if a dif-
ferential pressure measurement is used to control the flowrate out of the
reactor and the reactor has constant cross-sectional area. In this situation
ÏV
R
is specified as the additional equation.
3. Flowrate out of the reactor is specified.
This type of operation may be
achieved if the flowrate out of the reactor is controlled by a flow controller.
In this case
Q(t)
is specified. A semi-batch reactor is operated in this way
with
Q
=
0
until the reactor is filled with the reactants.
74
Nonconstant density – reactor operation
•
See the text for the derivation.
dV
R
dt
=
Q
f
P
j
f
j
c
jf
P
j
f
j
c
j
−
Q
+
P
i
∆
f
i
r
i
V
R
P
j
f
j
c
j
(50)
•
in which
f
j
is
f
j
=
∂f
∂c
j
•
and
∆
f
i
is
∆
f
i
=
X
j
ν
ij
f
j
=
X
j
ν
ij
∂f
∂c
j
which is a change in a derivative property upon reaction.
75
Nonconstant density – idea mixture
•
For the ideal mixture we have
f (c
j
)
=
P
j
c
j
V
â—¦
j
−
1
=
0
.
f
j
=
V
â—¦
j
the pure component specific volumes
•
The
∆
f
i
are given by
∆
f
i
=
∆
V
â—¦
i
the change in specific volume upon reaction
i
.
•
So the reactor volume can be calculated from
dV
R
dt
=
Q
f
−
Q
+
X
i
∆
V
â—¦
i
r
i
V
R
76
Nonconstant density
Unknowns (
ns
+
2
):
VR, Q, nj, j
=
1
, . . . , ns
Component balances:
dnj
dt
=
Qf cjf
−
Qcj
+
RjVR, j
=
1
, . . . , ns
Defined quantities:
nj
=
cjVR
Ï
=
P
j cjMj
∆
V
â—¦
i
=
P
j νijV
â—¦
j
(i) constant density:
Ï
=
Ï
0
(ii) ideal mixture:
P
j cjV
â—¦
j
=
1
1
.
vol
VR
=
VR
0
Q
=
Qf
VR
=
VR
0
Q
=
Qf
+
P
i
∆
V
â—¦
i
riVR
2
.
mass
VR
=
VR
0
Q
=
Qf
dVR
dt
=
Qf (
1
−
Ïf /Ï)
+
P
i
∆
V
â—¦
i
riVR
Q
=
Qf Ïf /Ï
3
. Q
dVR
dt
=
Qf
−
Q
Q
specified
dVR
dt
=
Qf
−
Q
+
P
i
∆
V
â—¦
i
riVR
Q
specified
Table 1: Reactor balances for constant-density and ideal-mixture assumptions.
77
Nonconstant density
Unknowns (
ns
+
2
):
VR, Q, nj, j
=
1
, . . . , ns
Component balances:
dnj
dt
=
Qf cjf
−
Qcj
+
RjVR, j
=
1
, . . . , ns
Defined quantities:
nj
=
cjVR
Ï
=
P
j cjMj
∆
fi
=
P
j νij
∂f
∂cj
Equation of state:
f (c
1
, c
2
, . . . , cns )
=
0
DAEs
ODEs
1
.
vol
VR
=
VR
0
f (cj)
=
0
VR
=
VR
0
Q
=
Qf
P
j fjcjf
P
j fjcj
+
P
i
∆
firiVR
P
j fjcj
2
.
mass
ÏVR
=
Ï
0
VR
0
f (cj)
=
0
dVR
dt
=
Qf
P
j fjcjf
P
j fjcj
−
Q
+
P
i
∆
firiVR
P
j fjcj
Q
=
Qf Ïf /Ï
3
. Q
Q
specified
f (cj)
=
0
dVR
dt
=
Qf
P
j fjcjf
P
j fjcj
−
Q
+
P
i
∆
firiVR
P
j fjcj
Q
specified
Table 2: Reactor balances for general equation of state.
78
Semi-batch polymerization
•
Consider a solution polymerization reaction, which can be modeled as a first-
order, irreversible reaction
M
k
-
→
P
r
=
kc
M
•
A 20 m
3
semi-batch reactor is initially charged with solvent and initiator to
half its total volume.
•
A pure monomer feed is slowly added at flowrate
Q
f
0
=
1
m
3
/min to fill the
reactor in semi-batch operation to control the heat release.
79
Semi-batch polymerization
•
Consider two cases for the subsequent reactor operation.
1. The monomer feed is shut off and the reaction goes to completion.
2. The monomer feed is adjusted to keep the reactor filled while the reaction
goes to completion.
•
Calculate the total polymer mass production, and the percentage increase in
polymer production achieved in the second operation.
80
The physical properties
•
You may assume an ideal mixture
•
The densities of monomer and polymer are
Ï
M
=
800
kg/m
3
Ï
P
=
1100
kg/m
3
•
The monomer molecular weight is
M
M
=
100
kg/kmol
•
The rate constant is
k
=
0
.
1
min
−
1
.
81
Semi-batch polymerization
•
While the reactor is filling, the monomer mole balance is
d(c
M
V
R
)
dt
=
Q
f
0
c
Mf
−
kc
M
V
R
in which
c
Mf
=
Ï
M
/M
M
is given, and
Q
f
=
Q
f
0
is constant during the filling
operation.
•
We denote the total number of moles of monomer by
M
=
c
M
V
R
, and can
write the monomer balance as
dM
dt
=
Q
f
0
c
Mf
−
kM
(51)
M(
0
)
=
0
82
Semi-batch polymerization
•
For an ideal mixture, the volume is given by
dV
R
dt
=
Q
f
0
+
∆
V kM
(52)
V
R
(
0
)
=
10
m
3
in which
∆
V
=
(
1
/Ï
P
−
1
/Ï
M
) M
M
83
The polymer mass production
•
To compute the polymer mass, we note from the stoichiometry that the mass
production rate of polymer
e
R
P
is
e
R
P
= −
R
M
M
M
•
The mass balance for total polymer
e
P
is given by
d
e
P
dt
=
e
R
p
V
R
=
kc
M
M
M
V
R
=
(kM
M
)M
(53)
84
Semi-batch polymerization
•
The text solves this problem analytically. Instead, let’s solve it numerically.
•
Let
t
1
be the time that the reactor fills.
•
We need an ODE solver that is smart enough to stop when the reactor fills,
because we do not know this time
t
1
. The ODE solver needs to find it for us.
•
dasrt
is an ODE solver with the added capability to find the time at which
some event of interest occurs.
85
Finding the time for filling the reactor
•
The ODE solver finds the time at which
V
R
=
20
m
3
t
1
=
11
.
2
min
•
Note the reactor would have filled in 10 min if the density were constant.
•
The extra time reflects the available volume created by converting some of
the monomer to polymer during filling.
•
After
t
1
we consider the two operations.
86
Operation 1.
•
In the first operation,
Q
f
=
0
after
t
1
.
•
Notice the reactor volume decreases after
t
1
because
∆
V
is negative.
87
Semi-batch polymerization
0
5
10
15
20
0
10
20
30
40
50
V
R
(
m
3
)
2
1
time (min)
Figure 4: Semi-batch reactor volume for primary monomer addition (operation
1) and primary plus secondary monomer additions (operation 2).
88
-0.2
0
0.2
0.4
0.6
0.8
1
0
10
20
30
40
50
Q
f
(
m
3
/
min
)
2
1
time (min)
Figure 5: Semi-batch reactor feed flowrate for primary monomer addition (op-
eration 1) and primary plus secondary monomer additions (operation 2).
89
0
1000
2000
3000
4000
5000
6000
0
10
20
30
40
50
monomer
(kg)
2
1
time (min)
Figure 6: Semi-batch reactor monomer content for primary monomer addition
(operation 1) and primary plus secondary monomer additions (operation 2).
90
0
2000
4000
6000
8000
10000
12000
0
10
20
30
40
50
polymer
(kg)
2
1
time (min)
Figure 7: Semi-batch reactor polymer content for primary monomer addition
(operation 1) and primary plus secondary monomer additions (operation 2).
91
Operation 2.
•
Because the reactor volume is constant, we can solve Equation 52 for the feed
flowrate during the secondary monomer addition
Q
f
= −
∆
V kM
•
Operation 2 is also shown in the figures.
•
Notice the final polymer production is larger in operation 2 because of the
extra monomer addition.
92
Polymer production rate
•
We can perform an independent, simple calculation of the total polymer in
operation 2. Useful for debugging the computation.
•
In operation 2, 10 m
3
of polymer are produced because in an ideal mixture,
the volumes are additive. Therefore
e
P
2
=
(V
R
−
V
R
0
)Ï
P
=
10
m
3
×
1100
kg/m
3
=
11000
kg
in good agreement with the long-time solution for operation 2.
•
The increase in production rate is
e
P
2
−
e
P
1
e
P
1
×
100%
=
22
.
5%
93
•
By using the volume of the reactor more efficiently, the total polymer produc-
tion increases 22.5%.
94
The Plug-Flow Reactor (PFR)
•
Plug flow in a tube is an ideal-flow assumption in which the fluid is well mixed
in the radial and angular directions.
•
The fluid velocity is assumed to be a function of only the axial position in the
tube.
•
Plug flow is often used to approximate fluid flow in tubes at high Reynolds
number. The turbulent flow mixes the fluid in the radial and angular direc-
tions.
•
Also in turbulent flow, the velocity profile is expected to be reasonably flat in
the radial direction except near the tube wall.
95
Thin Disk Volume Element
Given the plug-flow assumption, it is natural to take a thin disk for the reactor
volume element
Q
c
j
Q(z
+
∆
z)
c
j
(z
+
∆
z)
Q(z)
c
j
(z)
Q
f
c
jf
z
R
j
∆
V
z
+
∆
z
z
}|
{
96
Thin Disk Volume Element
•
Expressing the material balance for the volume element
∂
c
j
∆
V
∂t
=
c
j
Q
z
−
c
j
Q
z
+
∆
z
+
R
j
∆
V
(54)
•
Dividing the above equation by
∆
V
and taking the limit as
∆
V
goes to zero
yields,
∂c
j
∂t
| {z }
accumulation
= −
∂
c
j
Q
∂V
|
{z
}
convection
+
R
j
|{z}
reaction
(55)
97
Length or volume as independent variable
•
If the tube has constant cross section,
A
c
, then velocity,
v
, is related to vol-
umetric flowrate by
v
=
Q/A
c
, and axial length is related to tube volume by
z
=
V /A
c
,
•
Equation 55 can be rearranged to the familiar form [1, p.584]
∂c
j
∂t
= −
∂
c
j
v
∂z
+
R
j
(56)
98
Steady-State Operation
•
Setting the time derivative in Equation 55 to zero gives,
d(c
j
Q)
dV
=
R
j
(57)
•
The product
c
j
Q
=
N
j
is the total molar flow of component
j
. One also can
express the PFR mole balance in terms of the molar flow,
dN
j
dV
=
R
j
(58)
99
Volumetric Flowrate for Gas-Phase Reactions
•
To use Equation 58 for designing a gas-phase reactor, one has to be able to
relate the volumetric flowrate,
Q
, to the molar flows,
N
j
, j
=
1
,
2
, . . . , n
s
.
•
The important piece of information tying these quantities together is, again,
the equation of state for the reaction mixture,
f (T , P , c
j
)
=
0
.
•
Because the molar flow and concentration are simply related,
N
j
=
c
j
Q
(59)
the equation of state is also a relation between temperature, pressure, molar
flows, and volumetric flowrate.
100
Ideal Gas Equation of State
•
The ideal-gas equation of state,
c
=
P /RT
, can be stated in terms of molar
concentrations,
c
j
, as
X
j
c
j
=
P
RT
•
In terms of molar flows, the equation of state is
P
j
N
j
Q
=
P
RT
•
One can solve the previous equation for the volumetric flowrate,
Q
=
RT
P
X
j
N
j
(60)
101
Ideal Gas Equation of State
•
To evaluate the concentrations for use with the reaction rate expressions, one
simply rearranges Equation 59 to obtain
c
j
=
N
j
Q
=
P
RT
N
j
P
j
N
j
(61)
102
Volumetric Flowrate for Liquid-Phase Reactions
•
Consider the equation of state for a liquid-phase system to be arranged in
the form
Ï
=
f (T , P , c
j
)
•
The mass density is related to the volumetric flowrate and total mass flow,
M
=
P
j
N
j
M
j
, via
M
=
ÏQ
(62)
•
Multiplying Equation 58 by
M
j
and summing on
j
produces
dM
dV
=
0
,
M(
0
)
=
M
f
in which
M
f
is the feed mass flowrate.
•
The total mass flow in a PFR is constant.
103
Volumetric Flowrate for Liquid-Phase Reactions
•
We can solve for the volumetric flowrate by rearranging Equation 62
Q
=
M
f
Ï
(63)
•
If the liquid density is considered constant,
Ï
=
Ï
f
, then
Q
=
Q
f
,
constant density
(64)
and the volumetric flowrate is constant and equal to the feed value.
•
Equation 64 is used often for liquid-phase reactions.
104
Volumetric Flowrate for Liquid-Phase Reactions
•
If we denote the time spent in the tube by
Ï„
=
V /Q
, if
Q
is constant, we can
rewrite Equation 57 as
dc
j
dτ
=
R
j
,
constant flowrate
(65)
which is identical to the constant-volume batch reactor.
•
For the constant-flowrate case, the steady-state profile in a PFR starting from
a given feed condition is also the transient profile in a batch reactor starting
from the equivalent initial condition.
105
Single Reaction Systems – Changing flowrate in a PFR
•
A pure vapor stream of A is decomposed in a PFR to form B and C
A
k
-
→
B
+
C
(66)
•
Determine the length of 2.5 cm inner-diameter tube required to achieve
35% conversion of A. The reactor temperature is 518
â—¦
C
and the pressure is
2.0 atm. Assume the pressure drop is negligible.
•
The reaction rate is first order in A,
k
=
0
.
05
sec
−
1
at the reactor temperature,
and the feed flowrate is 35 L/min.
106
Changing flowrate in a PFR
•
The mole balance for component A gives
dN
A
dV
=
R
A
•
The production rate of A is
R
A
= −
r
= −
kc
A
.
•
Substituting the production rate into the above equation gives,
dN
A
dV
= −
kN
A
/Q
(67)
•
The volumetric flowrate is not constant, so we use Equation 60, which as-
sumes an ideal-gas equation of state,
Q
=
RT
P
(N
A
+
N
B
+
N
C
)
(68)
107
Changing flowrate in a PFR
•
The ideal-gas assumption is reasonable at this reactor temperature and pres-
sure.
•
One can relate the molar flows of B and C to A using the reaction stoichiom-
etry. The mole balances for B and C are
dN
B
dV
=
R
B
=
r
dN
C
dV
=
R
C
=
r
•
Adding the mole balance for A to those of B and C gives
d (N
A
+
N
B
)
dV
=
0
d (N
A
+
N
C
)
dV
=
0
•
The stoichiometry does not allow the molar flow
N
A
+
N
B
or
N
A
+
N
C
to change
with position in the tube.
108
Changing flowrate in a PFR
•
Because
N
A
+
N
B
and
N
B
+
N
C
are known at the tube entrance, one can relate
N
B
and
N
C
to
N
A
,
N
A
+
N
B
=
N
A
f
+
N
B
f
N
A
+
N
C
=
N
A
f
+
N
C
f
•
Rearranging the previous equations gives,
N
B
=
N
A
f
+
N
B
f
−
N
A
N
C
=
N
A
f
+
N
C
f
−
N
A
•
Substituting the relations for
N
B
and
N
C
into Equation 68 gives
Q
=
RT
P
2
N
A
f
+
N
B
f
+
N
C
f
−
N
A
109
Changing flowrate in a PFR
•
Because the feed stream is pure A,
N
B
f
=
N
C
f
=
0
, yielding
Q
=
RT
P
2
N
A
f
−
N
A
•
Substituting this expression in Equation 67 gives the final mole balance,
dN
A
dV
= −
k
P
RT
N
A
2
N
A
f
−
N
A
•
The above differential equation can be separated and integrated,
Z
N
A
N
Af
2
N
Af
−
N
A
N
A
dN
A
=
Z
V
0
−
kP
RT
dV
110
Changing flowrate in a PFR
•
Performing the integration gives,
2
N
Af
ln
N
A
/N
Af
+
N
Af
−
N
A
= −
kP
RT
V
•
The conversion of component
j
for a plug-flow reactor operating at steady
state is defined as
x
j
=
N
jf
−
N
j
N
jf
•
Because we are interested in the
V
corresponding to 35% conversion of A, we
substitute
N
A
=
(
1
−
x
A
)N
Af
into the previous equation and solve for V,
V
= −
RT
kP
N
Af
[
2 ln
(
1
−
x
A
)
+
x
A
]
111
Changing flowrate in a PFR
•
Because
Q
f
=
N
A
f
RT /P
is given in the problem statement and the tube
length is desired, it is convenient to rearrange the previous equation to obtain
z
= −
Q
f
kA
c
[
2 ln
(
1
−
x
A
)
+
x
A
]
•
Substituting in the known values gives
z
= −
35
×
10
3
cm
3
/
min
0
.
05 sec
−
1
60 sec
/
min
!
4
Ï€ (
2
.
5 cm
)
2
[
2 ln
(
1
−
.
35
)
+
.
35
]
z
=
1216 cm
=
12
.
2 m
112
Multiple-Reaction Systems
•
The modeler has some freedom in setting up the material balances for a plug-
flow reactor with several reactions.
•
The most straightforward method is to write the material balance relation for
every component,
dN
j
dV
=
R
j
,
j
=
1
,
2
, . . . , n
s
R
j
=
n
r
X
i
=
1
ν
ij
r
i
,
j
=
1
,
2
, . . . , n
s
•
The reaction rates are expressed in terms of the species concentrations.
•
The
c
j
are calculated from the molar flows with Equation 59
•
Q
is calculated from Equation 60, if an ideal-gas mixture is assumed.
113
Benzene pyrolysis in a PFR
•
Hougen and Watson [3] analyzed the rate data for the pyrolysis of benzene
by the following two reactions.
•
Diphenyl is produced by the dehydrogenation of benzene,
2C
6
H
6
k
1
-
*
)
-
k
−
1
C
12
H
10
+
H
2
2B
-
*
)
-
D
+
H
114
Benzene pyrolysis in a PFR
•
Triphenyl is formed by the secondary reaction,
C
6
H
6
+
C
12
H
10
k
2
-
*
)
-
k
−
2
C
18
H
14
+
H
2
(69)
B
+
D
-
*
)
-
T
+
H
(70)
•
The reactions are assumed to be elementary so that the rate expressions are
r
1
=
k
1
c
2
B
−
c
D
c
H
K
1
(71)
r
2
=
k
2
c
B
c
D
−
c
T
c
H
K
2
(72)
115
Benzene pyrolysis in a PFR
•
Calculate the tube volume required to reach 50% total conversion of the ben-
zene for a 60 kmol/hr feed stream of pure benzene.
•
The reactor operates at 1033
K
and 1.0 atm.
•
Plot the mole fractions of the four components versus reactor volume.
116
Benzene pyrolysis in a PFR
•
The rate and equilibrium constants at
T
=
1033K
and
P
=
1
.
0
atm are given
in Hougen and Watson,
k
1
=
7
×
10
5
L/mol
·
hr
k
2
=
4
×
10
5
L/mol
·
hr
K
1
=
0
.
31
K
2
=
0
.
48
117
Benzene pyrolysis in a PFR
•
The mole balances for the four components follow from the stoichiometry,
dN
B
dV
= −
2
r
1
−
r
2
(73)
dN
D
dV
=
r
1
−
r
2
(74)
dN
H
dV
=
r
1
+
r
2
(75)
dN
T
dV
=
r
2
(76)
•
The initial condition for the ODEs are
N
B
(
0
)
=
N
Bf
and
N
D
(
0
)
=
N
H
(
0
)
=
N
T
(
0
)
=
0
.
118
Benzene pyrolysis in a PFR
•
The total molar flux does not change with reactor volume.
Q
=
RT
P
N
Bf
(77)
•
The rate expressions are substituted into the four ODEs and they are solved
numerically.
•
The total conversion of benzene,
x
B
=
(N
Bf
−
N
B
)/N
Bf
, is plotted versus
reactor volume in Figure 8.
•
A reactor volume of 404 L is required to reach 50% conversion. The compo-
sition of the reactor versus reactor volume is plotted in Figure 9.
119
0
0.1
0.2
0.3
0.4
0.5
0.6
0
200
400
600
800
1000 1200 1400 1600
xB
V
(L)
Figure 8: Benzene conversion versus reactor volume.
120
0
0.2
0.4
0.6
0.8
1
0
200
400
600
800
1000 1200 1400 1600
yB
yH
yT
yD
yj
V
(L)
Figure 9: Component mole fractions versus reactor volume.
121
Ethane pyrolysis in the presence of NO
•
See the text for another worked PFR example.
122
Some PFR-CSTR Comparisons
•
We have two continuous reactors in this chapter: the CSTR and the PFR.
•
Let’s compare their steady-state efficiencies in converting reactants to prod-
ucts.
•
For simplicity, consider a constant-density, liquid
A
k
-
→
B
r
=
kc
n
A
(78)
•
For this situation, the steady-state PFR material balance is given by Equa-
tion 65
dc
A
dτ
= −
r (c
A
)
123
Some PFR-CSTR Comparisons
•
We rearrange and solve for the time required to change from the feed condi-
tion
c
Af
to some exit concentration
c
A
Ï„
=
Z
c
A
c
Af
1
r (c
0
A
)
dc
0
A
•
The area under the curve
1
/r (c
0
A
)
is the total time required to achieve the
desired concentration change.
1
r (c
0
A
)
c
Af
c
A
c
0
A
PFR
CSTR
124
Some PFR-CSTR Comparisons
•
To achieve this same concentration change in the CSTR, we start with Equa-
tion 37, and solve for
Ï„
giving
Ï„
=
c
Af
−
c
A
r (c
A
)
•
This result also can be interpreted as an area. Notice that this area is the
height,
1
/r (c
A
)
, times the width,
c
Af
−
c
A
, of the rectangle.
1
r (c
0
A
)
c
Af
c
A
c
0
A
PFR
CSTR
125
Some PFR-CSTR Comparisons
•
If
1
/r (c
A
)
is a
decreasing
function of
c
A
, or, equivalently,
r (c
A
)
is an
increas-
ing
function of
c
A
, to achieve the same conversion, the PFR time (or volume,
V
R
=
Q
f
Ï„
) is less than the CSTR time (volume).
•
The PFR reaction rate varies with length. The rate is high at the entrance to the
tube where the concentration of A is equal to the feed value, and decreases
with length as the concentration drops. At the exit of the PFR, the rate is the
lowest of any location in the tube.
•
Now considering that the entire volume of the CSTR is reacting at this
lowest
rate of the PFR, it is intuitively obvious that more volume is required for the
CSTR to achieve the same conversion as the PFR.
126
Some PFR-CSTR Comparisons
•
If the reaction order is positive (the usual case), the PFR is more efficient. If
the reaction order is negative, the CSTR is more efficient.
1
r (c
0
A
)
c
Af
c
A
c
0
A
CSTR
PFR
127
The PFR versus CSTR with separation
•
The PFR achieves higher conversion than an equivalent volume CSTR for the
irreversible reaction with first-order kinetics
A
-
→
B
r
=
kc
A
•
Consider the case in which we add separation.
•
Find a single CSTR and separator combination that achieves the same con-
version as the PFR.
128
The PFR versus CSTR with separation
•
The issue is to increase the CSTR achievable conversion using separation.
•
Education in chemical engineering principles leads one immediately to con-
sider recycle of the unreacted A as a means to increase this conversion.
129
The PFR versus CSTR with separation
N
A
0
N
A
1
N
A
2
N
A
N
A
0
N
A
αN
A
2
αN
A
2
pure B
V
R
V
R
•
In the text, we show how to find the recycle flowrate so this system achieves
the PFR conversion.
130
CSTR Equivalence Principle.
•
This example was motivated by a recent result of Feinberg and Ellison called
the CSTR Equivalence Principle of Reactor-Separator Systems [2].
•
This surprising principle states:
For a given reaction network with
n
i
linearly independent reactions, any
steady state that is achievable by any reactor-separator design with total
reactor volume
V
is achievable by a design with
not more than
n
i
+
1
CSTRs
, also of total reactor volume
V
. Moreover the concentrations,
temperatures and pressures in the CSTRs are arbitrarily close to those
occurring in the reactors of the original design.
131
Stochastic Simulation of Chemical Reactions
•
We wish to introduce next a topic of increasing importance to chemical engi-
neers, stochastic (random) simulation.
•
In stochastic models we simulate quite directly the random nature of the
molecules.
•
We will see that the deterministic rate laws and material balances presented in
the previous sections can be captured in the stochastic approach by allowing
the numbers of molecules in the simulation to become large.
•
The stochastic modeling approach is appropriate if the random nature of the
system is one of the important features to be captured in the model.
•
These situations are becoming increasingly important to chemical engineers
as we explore reactions at smaller and smaller length scales.
132
Stochastic Simulation of Chemical Reactions
•
For example, if we are modeling the chemical transformation by reaction of
only a few hundreds or thousands of molecules at an interface, we may want
to examine explicitly the random fluctuations taking place.
•
In biological problems, we often consider the interactions of only several hun-
dred or several thousand protein molecules and cells.
•
In sterilization problems, we may wish to model the transient behavior until
every last organism is eliminated.
133
Stochastic Simulation of Chemical Reactions
•
Assume we have only a hundred molecules moving randomly in the gas phase
A
k
1
-
→
B
(79)
B
k
2
-
→
C
(80)
in a constant-volume batch reactor.
•
The
probability
of reaction is assumed proportional to the
r
1
=
k
1
x
A
r
2
=
k
2
x
B
in which
x
j
is the
number
of component
j
molecules in the reactor volume.
•
Note
x
j
is an integer, unlike the deterministic model’s
c
j
, which is real.
134
Stochastic Simulation of Chemical Reactions
•
The basic idea of the Gillespie algorithm is to: (i) choose randomly the time
at which the next reaction occurs, and (ii) choose randomly which reactions
occurs at that time.
1. Initialize. Set integer counter
n
to zero. Set the initial species numbers,
x
j
(
0
), j
=
1
, . . . n
s
. Determine stoichiometric matrix
ν
and reaction probabil-
ity laws (rate expressions)
r
i
=
k
i
h(x
j
)
for all reactions.
2. Compute reaction probabilities,
r
i
=
k
i
h(x
j
)
. Compute total reaction proba-
bility
r
tot
=
P
i
r
i
.
135
Stochastic Simulation of Chemical Reactions
3. Select two random numbers,
p
1
, p
2
, from a uniform distribution on the inter-
val
(
0
,
1
)
. Let the time interval until the next reaction be
e
t
= −
ln
(p
1
)/r
tot
(81)
136
Stochastic Simulation of Chemical Reactions
4. Determine reaction
m
to take place in this time interval.
The idea
here is to partition the interval (0,1) by the relative sizes of each
reaction probability and then “throw a dart†at the interval to pick
the reaction that occurs.
In this manner,
all reactions are pos-
sible,
but the reaction is selected in accord with its probability.
p
2
r
1
r
1
+
r
2
r
2
r
1
+
r
2
0
1
137
Stochastic Simulation of Chemical Reactions
5. Update the simulation time
t(n
+
1
)
=
t(n)
+
e
t
. Update the species numbers
for the single occurrence of the
m
th reaction via
x
j
(n
+
1
)
=
x
j
(n)
+
ν
mj
,
j
=
1
, . . . n
s
Let
n
=
n
+
1
. Return to Step 2.
138
Stochastic Simulation of Chemical Reactions
•
If
r
tot
is the total probability for reaction,
e
−
r
tot
e
t
is the probability that a re-
action has not occurred during time interval
e
t
.
•
We will derive this fact in Chapter 8 when we develop the residence-time
distribution for a CSTR.
•
The next figure shows the results of this algorithm when starting with
x
A
=
100
molecules.
139
Stochastic Simulation of Chemical Reactions
0
20
40
60
80
100
0
1
2
3
4
5
xA
xB
xC
xj
t
140
Stochastic Simulation of Chemical Reactions
•
Notice the random aspect of the simulation gives a rough appearance to the
number of molecules versus time, which is quite unlike any of the determin-
istic simulations.
•
Because the
number
of molecules is an integer, the simulation is actually
discontinuous with jumps between simulation times.
•
But in spite of the roughness, we already can make out the classic behavior
of the series reaction: loss of starting material A, appearance and then dis-
appearance of the intermediate species B, and slow increase in final product
C.
141
Stochastic Simulation of Chemical Reactions
•
Next we explore the effect of increasing the initial number of A molecules on
a single simulation. The results for 1000 and 4000 initial A molecules are
shown in the next figures.
142
Stochastic Simulation of Chemical Reactions
0
200
400
600
800
1000
0
1
2
3
4
5
xA
xB
xC
xj
t
143
Stochastic Simulation of Chemical Reactions
0
800
1600
2400
3200
4000
0
1
2
3
4
5
xA
xB
xC
xj
t
144
Stochastic Simulation of Chemical Reactions
•
We see the random fluctuations become less pronounced. Notice that even
with only 4000 starting molecules, the results compare very favorably with
the deterministic simulation shown previously.
•
Another striking feature of the stochastic approach is the trivial level of pro-
gramming effort required to make the simulations.
•
The biggest numerical challenge is producing the pseudorandom numbers
and many well-developed algorithms are available for that task.
•
The computational time required for performing the stochastic simulation
may, however, be large.
145
Hepatitis B virus modeling
146
Hepatitis B virus modeling
nucleotides
cccDNA
−
−
−
−
−
−→
rcDNA
(82)
nucleotides
+
rcDNA
−
−
−
−
−
−→
cccDNA
(83)
amino acids
cccDNA
−
−
−
−
−
−→
envelope
(84)
cccDNA
−
−
−
−
−
−→
degraded
(85)
envelope
−
−
−
−
−
−→
secreted or degraded
(86)
rcDNA
+
envelope
−
−
−
−
−
−→
secreted virus
(87)
147
Hepatitis B virus modeling
•
The reaction rates and production rates for Reactions 82–87 are given by











r
1
r
2
r
3
r
4
r
5
r
6











=











k
1
x
A
k
2
x
B
k
3
x
A
k
4
x
A
k
5
x
C
k
6
x
B
x
C















R
A
R
B
R
C




=




r
2
−
r
4
r
1
−
r
2
−
r
6
r
3
−
r
5
−
r
6




(88)
in which A is cccDNA, B is rcDNA, and C is envelope.
148
Hepatitis B virus modeling
•
Assume the systems starts with a single cccDNA molecule and no rcDNA and
no envelope protein, and the following rate constants
h
x
A
x
B
x
C
i
T
=
h
1
0
0
i
T
(89)
k
T
=
h
1
0
.
025
1000
0
.
25
2
7
.
5
×
10
−
6
i
(90)
149
Average stochastic is not deterministic.
0
5
10
15
20
25
0
50
100
150
200
cccDNA
deterministic
average stochastic
t
(days)
0
50
100
150
200
250
0
50
100
150
200
rcDNA
deterministic
average stochastic
t
(days)
150
0
2000
4000
6000
8000
10000
0
50
100
150
200
envelope
deterministic
average stochastic
t
(days)
151
Hepatitis B virus modeling
0
5
10
15
20
25
30
35
0
50
100
150
200
cccDNA
deterministic
stochastic 1
stochastic 2
t
(days)
Figure 10: Species cccDNA versus time for hepatitis B virus model; two repre-
sentative stochastic trajectories.
152
Hepatitis B virus modeling
•
The simulation of the deterministic model and an average of 500 stochastic
simulations are
not
the same.
•
Figure 10 shows
two
representative stochastic simulations for only the cc-
cDNA species.
•
Notice the first stochastic simulation does fluctuate around the deterministic
simulation as expected.
•
The second stochastic simulation, however, shows complete extinction of the
virus. That is another possible steady state for the stochastic model.
153
Average stochastic is not deterministic.
•
In fact, it occurs for 125 of the 500 simulations. So the
average
stochastic
simulation consists of 75% trajectories that fluctuate about the deterministic
trajectory and 25% trajectories that go to zero.
0
5
10
15
20
25
0
50
100
150
200
cccDNA
deterministic
average stochastic
t
(days)
154
Summary
•
We have introduced four main reactor types in this chapter.
•
the batch reactor
•
the continuous-stirred-tank reactor (CSTR)
•
the semi-batch reactor
•
the plug-flow reactor (PFR).
155
Summary
BATCH
d(cjVR)
dt
=
RjVR
(91)
constant volume
dcj
dt
=
Rj
(92)
CSTR
d(cjVR)
dt
=
Qf cjf
−
Qcj
+
RjVR
(93)
constant density
dcj
dt
=
1
Ï„
(cjf
−
cj)
+
Rj
(94)
steady state
cj
=
cjf
+
Rjτ
(95)
SEMI-BATCH
d(cjVR)
dt
=
Qf cjf
+
RjVR
(96)
PFR
∂cj
∂t
=
−
∂(cjQ)
∂V
+
Rj
(97)
steady state
d(cjQ)
dV
=
Rj
(98)
constant flowrate
dcj
dτ
=
Rj,
Ï„
=
V /Qf
(99)
156
Summary
We also have introduced some of the basic reaction-rate expressions.
•
first order, irreversible
•
first order, reversible
•
second order, irreversible
•
n
th order, irreversible
•
two first-order reactions in series
•
two first-order reactions in parallel
•
two second-order, reversible reactions
157
Summary
•
We developed the equations required to compute the volume of the reactor if
there is a significant volume change upon reaction. We require an equation
of state for this purpose.
•
Several of these simple mass balances with basic rate expressions were solved
analytically.
•
In the case of multiple reactions with nonlinear rate expressions (i.e., not
first-order reaction rates), the balances must be solved numerically.
•
A high-quality ordinary differential equation (ODE) solver is indispensable for
solving these problems.
158
Summary
•
We showed that the PFR achieves higher conversion than the CSTR of the
same volume if the reaction rate is an increasing function of a component
composition (
n >
0
for an
n
th-order rate expression).
•
Conversely, the CSTR achieves higher conversion than the same-volume PFR
if the rate is a decreasing function of a component composition (
n <
0
).
•
Finally, we introduced stochastic simulation to model chemical reactions oc-
curring with
small
numbers of molecules.
159
Summary
•
The stochastic model uses basic probability to compute reaction rate. A given
reaction’s probability of occurrence is assumed proportional to the number
of possible combinations of reactants for the given stoichiometry.
•
Two pseudorandom numbers are chosen to determine: (i) the time of the
next reaction and (ii) the reaction that occurs at that time.
•
The smooth behavior of the macroscopic ODE models is recovered by the
random simulations in the limit of large numbers of reacting molecules.
•
With small numbers of molecules, however, the average of the stochastic
simulation does not have to be equal to the deterministic simulation. We
demonstrated this fact with the simple, nonlinear hepatitis B virus model.
160
References
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Transport Phenomena
. John
Wiley & Sons, New York, second edition, 2002.
[2] M. Feinberg and P. Ellison. General kinetic bounds on productivity and se-
lectivity in reactor-separator systems of arbitrary design: I. Principles.
Ind.
Eng. Chem. Res.
, 40(14):3181–3194, 2001.
[3] O. A. Hougen and K. M. Watson.
Chemical Process Principles. Part Three:
Kinetics and Catalysis
. John Wiley & Sons, New York, 1947.
161