Chemical Reaction Engineering
The rate of a reaction 𝐫𝐀
Since chemical reaction engineering concerns itself with the kinetics of the reaction, we must first define what is meant by
the “rate of a reaction”.
We define the rate of a reaction for component A as follows: “the amount of moles of A disappearing or being generated
per unit time per unit volume is the rate of the reaction, denoted as 𝐫𝐀 ”.
Equation for Rate
The rate of a reaction is determined by the rate law. For example, for the reaction,
A→B
The rate law could look like,
−r = k[A]
Here, [A] means the concentration of A in mol/L. We will use C to denote the concentration from now on. So,
−r = kC
In this case we say that the reaction is first order in A, as the exponent of C is 1. The rate law could also look like,
−r = kC
In this case, the reaction would be second order in A.
The rate law could also look something like this,
−r =
k C
1+k C
In which case the order would be hard to determine.
Rate for components of a reaction
Given the reaction,
aA + bB → cC + dD
The rate laws for each component are related as,
1
1
1
1
r = r = r = r
a
b
c
d
General Mole Balance for a System / Reactor
We now consider a system volume with a species j as shown below:
The mole balance on this system is,
[moles of j in per unit time] − [moles of j out per unit time] + [moles of j generated per unit time]
= [moles of j accumulated per unit time]
F −F +G =
dN
dt
If all the properties of a system, such as catalysis concentration, temperature, concentration of species are uniform
throughout the volume, then we can write the G term simply as the product of r and the system volume V.
F −F +rV=
dN
dt
If the system is not uniform, then r will vary over volume elements, and we will have to integrate it over space.
F −F +
r dV =
dN
dt
The Batch Reactor (BR)
A batch reactor is the simplest form of reactor to analyse. It is a closed reactor where the components are put in and allowed
to react until desired or equilibrium conditions are reached. Then, the products are taken out after sufficient time has passed.
For the batch reaction, since the system is closed, we have:
F = 0; F = 0
The equation for the reactor thus becomes,
r dV =
dN
dt
If the reactor is uniformly mixed, we will just have,
dN
=rV
dt
The Continuous Stirred Tank Reactor (CSTR)
The CSTR is a type of reactor which is operated at steady state. Components are being constantly placed in and the products
are being taken out. The reactor is perfectly mixed, as a result there is no time dependence, or position dependence on
temperature, concentration, and other physical properties within the reactor. The CSTR is usually used in liquid phase
reactions. We write the general mole balance equation:
F −F +
We note that as the reactor is at steady state, we have:
r dV =
dN
dt
= 0. And ∫ r dV = r V.
We thus have,
F −F +rV=0
𝐕=
𝐅𝐣𝟎 − 𝐅𝐣
−𝐫𝐣
This is the design equation for the CSTR.
The CSTR design equation thus gives us the required volume to reduce an incoming flow rate F to F when the rate of the
reaction is r . The molar flow rate F can just be written as:
F =C ⋅v
Here v is the volumetric flow rate.
We can rewrite the CSTR in terms of input and output concentration and volumetric flow rates,
V=
v C
−v C
r
Do note that the ideal CSTR design equation is not a differential equation but an algebraic one.
The Plug Flow Reactor (PFR) or the Tubular Reactor
The tubular reactor also called the PFR is a cylindrical pipe along with reaction takes place as the reactants flow down the
length of the reactor, normally operated at steady state. Since the concentration of the components vary in the axial
direction, the rate law also changes (unless the reaction is zero order). For our analysis we will consider the velocity to be
uniform across a very thin cross section, such that there is no variation in rate law along the radial direction, only the axial.
Let us do a mole balance on a sufficiently small section of the reactor, which we have shown as a cylinder within the
cylinder. The section is sufficiently small such that there is no variation in properties along it, so we can write the generation
term as simply r ΔV than an integral.
F| −F|
+ r ΔV =
dN
dt
We note that as the PFR reactor is operated at steady state, the section will too be at steady state and thus
F| −F|
= 0.
= −r ΔV
We get,
F|
−F|
=r
ΔV
Making the section volume tend to 0,
lim
→
F|
−F|
=r
ΔV
We notice that the left-hand side is basically the derivative of F with respect to V.
dF
=r
dV
This is the equation of the plug flow reactor. Understand that this is basically telling us, that the rate of change of the
molar outflow with respect to a change in the volume is simply the rate of the reaction.
We now ask, what reactor volume in a PFR reactor is required to reduce the incoming molar flow of A from F
write the equation in terms of A:
dV =
dF
r
to F , we
Integrating,
dF
=
r
V=
dF
−r
Question. Derive an expression of concentration (C ) over volume for the reaction A → B, given that it is first order in A
and is done inside a PFR reactor and the volumetric flow rate is constant at v
We have the rate law,
r = −kC
And,
F =C ⋅v
Where, v is the volumetric flow rate.
We can write the PFR design equation as,
dV =
−d(C ⋅ v )
kC
dV = −
v dC
⋅
k C
Integrating,
dV = −
v
k
dC
v
C
= − ⋅ ln
C
k
C
V=−
v
C
ln
k
C
C =C e
Conversion
Consider the reaction,
aA + bB → cC + dD
We divide by a, to make A the limiting reactant,
b
c
d
A+ B→ C+ D
a
a
a
Now, we ask the question, “how many moles of C are formed per mole of A?”, or “how many moles of B are used per mole
of A used?”. These questions will be answered by defining a quantity called the conversion factor, which is useful for
telling us the extent of reaction. We thus define,
X=
moles of A reacted
moles of A fed
The maximum conversion factor for a total irreversible reaction is 1.0, where all of A fed will have reacted. For reversible
reactions, the maximum conversion factor is X , that is the conversion factor to equilibrium concentration.
Conversion for Batch Reactor
For Batch Reactor we have the equation,
−
dN
= −r V
dt
N is the number of moles of A at time t. Let the conversion factor at time t be X , then N = N
equation becomes,
dN
d(N
=
dt
For a BR, N
− N X ) dN
d(N X )
=
−
dt
dt
dt
is constant, as once the reactants are fed, the reactor is closed. Thus,
N
⋅
dX
= −r V
dt
⋅ (1 − X ). So, the
This is the design equation for a batch reactor in terms of conversion.
We can calculate the time needed to achieve a certain conversion,
dX
−r V
t=N
Conversion for CSTR
For CSTR, we have a constant molar flow of F into the system. As the reactor is at steady state, at any point in time, we
can be guaranteed, that the molar rate at which the reactants are reacting is given by F ⋅ X at any point in time.
F X=
moles of A in moles of A reacted moles of A reacted
⋅
=
time
moles of A in
time
Thus, the molar flow rate at which unreacted A leaves the system is given by,
F = F (1 − X)
Recall that the entering molar flow rate is just the product of the concentration and the volume flow rate,
F
=C v
The concentration for gases can be calculated using the ideal gas law with C
pressure, y
=
=
(here P
is the partial gas
is the mole fraction of gas in the air, P is the partial pressure, R is the gas constant and T is the temperature)
Recall the equation for the CSTR,
V=
F
−F
−r
We substitute, F = F (1 − X), to yield:
V=
F X
−r
As the reactor is uniform all throughout, we can calculate the rate at exit since it is easy to measure the concentrations at the
reactor exit,
V=
F X
(−r )
Conversion for PFR
We model the PFR as having no concentration gradient along the radial direction. The mole balance equation of the PFR is,
−
dF
= −r
dV
We have,
F =F
− XF
dF = −F dX
F dX
= −r
dV
V=
F
⋅
dX
r
Rate as a function of Conversion
The first order dependence of reactor rate and conversion is given by,
−r = kC (1 − X)
This will be useful later.
Reactor Sizing
Reactor sizing involves one of two things, either determining a volume require for a certain amount of conversion or to
determine the conversion for a certain reactor volume.
We consider a reaction,
A→B
And go to the laboratory to determine the rate r of the reaction with respect to the conversion. A table is shown below.
Recalling the CSTR and PFR design equations, V =
instead of r . So, we calculate the values of
Next, we plot
and
and V = ∫
, we realize that we will need the value of
.
against X to achieve the following plot:
This we call the Levenspiel Plot. It will help us determine the volume required for CSTR and PFR.
Levenspiel Plot – Volume for CSTR
Since the Levenspiel plot is
against X, and the volume of a CSTR is simply,
X, the volume required for a CSTR is
simply the area of the given square:
Levenspiel Plot – Volume for PFR
The PFR design equation involves the integral, ∫
dX, therefore it would be the area under the curve.
It should be noted, V
,
>V
,
, always.
CSTR in Series
Let us consider two CSTR in series. The entry flow rate of the first CSTR is F , while the exit flow rate is F , it follows
that the entry flow rate of the second CSTR is F . Consider the conversions in the first reactor to be X and in the second to
be X . We have,
V =
F X
−r
And,
V =
If we had n CSTR in series, the volume of the n
reactor would be,
V =
Note that X here is defined as
F (X − X )
−r
F (X − X
−r
𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐀 𝐫𝐞𝐚𝐜𝐭𝐞𝐝
𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐀 𝐟𝐞𝐝 𝐢𝐧 𝐭𝐡𝐞 𝐟𝐢𝐫𝐬𝐭 𝐫𝐞𝐚𝐜𝐭𝐨𝐫
Recall that for the second CSTR, r
)
!
is measured at the exit (X ), and then the volume is calculated using (X − X ).
Approximating a PFR reactor using many CSTRs in series
Recall that,
V =
F (X − X
−r
)
Suppose we have a theoretically infinite number of reactors, in that case, the reactor volume would be very small, as would
the difference between the conversions in two adjacent reactors. In that case V → dV and X − X
→ dX.
We would get,
dV =
F
−r
dX
This is the design equation for a PFR! In essence, an ideal PFR reactor is basically a theoretically infinite chain of ideal
CSTR reactors.
This can be looked at graphically,
As we are increasing the number of CSTR reactors, our area describing the CSTR volume is slowly approaching the value
of the integral, which is the PFR volume.
PFRs in Series
We have the equation for the PFR,
F
dX
−r
V=
If we had two PFRs, the first taking the conversion to X and the second to X , we would have,
V=
Notice that this integral sum is simply ∫
F
dX +
−r
F
dX
−r
dX. Thus unlike CSTRs, we do not get any advantage when we combine
PFRs in series to achieve a specific conversion. We could just have a reactor with a volume V + V .
Space Time
We define a quantity called space time (τ) as,
τ=
V
v
Here, V is the volume of the reactor, and v is the volumetric flow rate of the reactor. What does this quantity represent? It
represents the amount of time it takes for the old CSTR constituents to be completely replaced with new constituents.
The Full Rate Law of a Reaction
The rate law can be written as,
−r = [k (T)][f(C , C , … )]
Some examples of first and second order rate laws are shown,
Non-Elementary Rate Laws
For the reaction,
CO + Cl → COCl
The rate law is,
−r
=k⋅C
⋅C
.
The reaction is first order w.r.t to CO, 1.5 order wrt to Cl and 2.5 order overall. However, some rate laws cannot be
reduced to an order, such as the reaction:
2N O → 2N + O
The rate law expression for this reaction is,
−r
k
C
1+k C
=
Rate Laws for Reversible Reactions
All rate laws for reversible reactions must correspond to thermodynamically relating the reacting species concentrations to
the equilibrium state.
For the reaction,
aA + bB ⎯⎯⎯ cC + dD
The equilibrium constant is written as,
K =
C
,
C
,
C
,
C
,
Now, consider an example reaction, the dimerization of benzene to diphenyl,
2C H
⎯⎯ C H
+H
Or symbolically as,
2B → D + H
We define the forward reaction coefficient 𝐤 𝐁 and the backward reaction coefficient 𝐤
𝐁
with respect to benzene.
Benzene is thus being depleted at the rate of r .
r
= −k C
However, there is a reverse reaction between hydrogen and diphenyl at equilibrium that creates benzene, thus the reverse
rate of reaction is,
r
=k
C C
The total rate of formation of benzene is given by,
r =r
Recall that the constant
equilibrium constant K .
Thus,
+r
= −k C + k
C C
− k (C +
C C
k
k
, that is ratio between forward reaction coefficient and backward reaction coefficient is the
−r = −k
C +
C C
K
Using the mole relations from the reaction, we can obtain the rates for other components, for example for diphenyl
formation,
r =
C C
1
k
r =+
C +
−2
2
K
Temperature Dependence on Equilibrium Constant (Van’t Hoff Equation)
The equilibrium constant K (T) at temperature T given that the equilibrium constant at T is K (T ), given by the Van’t
Hoff equation:
K (T) = K (T )e
Here,
ΔH
− The heat of the reaction at standard state (S.T.P)
R − The gas constant
Temperature Dependence on Rate Constant (Arrhenius Equation)
The rate constant k at temperature T is given by the equation,
k(T) = Ae
Here, E is the activation energy of the reaction. A is the Arrhenius factor. Usually, we do not need to calculate A, as a
reference rate constant k at a certain temperature T is given and we can find out the rate constant k at a given temperature
T by simply taking the ratio of the equations and cancelling A. Thus, this equation is more useful,
k(T) = k(T )e
Stoichiometric Balance for Batch System
Consider a batch reactor undergoing the reaction,
b
c
d
A+ B+I→ C+ D+I
a
a
a
(I being the inert component here)
At t = 0, we have, N , N , N , N , N and at t = t, we have N , N , N , N , and N .
Since the reaction is done at the basis of A, we define the conversion factor X in terms of A.
N = N (1 − X) = N
− XN
Now we have,
b
XN
a
c
ΔN = +
XN
a
ΔN = −
N =N
N =N
b
− XN
a
c
+ XN
a
ΔN = +
d
XN
a
N =N
d
+ XN
a
This is relatively easy to understand. For every mole of A consumed, moles of B are consumed, moles of C are
generated, and moles of D are generated.
We will define another quantity δ, that will tell us about the change in moles of the overall reaction.
δ=
change in total no. of moles
= ΔN + ΔN + ΔN + ΔN
moles of A reacted
δ=
d c b
+ − −1
a a a
Thus, the equation for total number of moles (N ) becomes,
N =N
+ δ(N X)
To find concentration, we can simply divide the moles by the volume,
C =
N
N
=
V
N
N
C =
=
V
C =
N
N
=
V
N
N
C =
=
V
−N X
V
b
−
XN
a
V
c
+
XN
a
V
+
d
XN
a
V
Since the batch reactor is almost always at constant volume, we can replace the V with V and
as C .
C = C (1 − X)
C =C
−
b
C X=C
a
C
C
b
− X
a
C =C
+
c
C X=C
a
C
C
c
+ X
a
C =C
+
d
C X=C
a
C
C
d
+ X
a
We define a parameter denoted as Θ that is,
Θ =
moles of species 'i' initially N
=
moles of A initially
N
=
C
C
=
y
y
The equations for concentrations become simplified:
C =C
C =C
C =C
b
Θ − X
a
c
Θ + X
a
d
Θ + X
a
Now that we have the concentrations in terms of conversion, we can easily write the rate law, if the reaction is first order in
A and B,
b
−r = kC C = kC (1 − X) Θ − X
a
Stoichiometric Balance for Flow System
The stoichiometric balance for a flow system as shown is virtually identical to the batch system, except we now replace the
initial number of moles (N ) with the molar flow rate (F ).
F = F (1 − X)
b
Θ − X
a
c
Θ + X
a
F =F
F =F
d
Θ + X
a
F =F
Here,
Θ =
F
F
Similarly, the concentration can be calculated as,
C =
F
v
Here v is the volumetric flow rate.
For liquids, unless phase change is taking place, the volumes are practically constant, so we can simply use v to calculate
the concentration.
C =
F
v
For gas phases, we will have a bit of a problem since the gas volume may change with the reaction. An example is the
reaction,
N + 3H → 2NH
Here the reaction will cause a decrease in the number of moles. In this case, the molar flow rate across the reactor will be
changing as the reaction progresses. And because equal moles occupy equal volume, the volumetric flow rate will be
changing as well. Thus, the total concentration, C at any point in the reactor will be written as:
C =
F
P
=
v
ZRT
Here Z is the compressibility factor of the gas as studied while dealing with real gas systems. If the gas is assumed to be
ideal, Z = 1.
At reactor entrance,
C
=
F
P
=
v
Z RT
Taking the ratio, and assuming negligible changes in the compressibility factor Z ≈ Z , we have,
v=v
F
F
P
P
T
T
We can now express, C as,
C =
Know that
F
=
v v
F
F
F
P
P
T
T
=
F
F
F
v
P
P
T
T
=C .
C =C
F
F
P
P
T
T
Finally, note that, F is just the sum of moles, F = F + F + F + ⋯, thus the ratio,
Thus, we have,
C =C y
P
P
T
T
Now, let us expression the concentration for gas flow systems in terms of conversion.
is just the mole fraction of j.
Recall,
F =F
+ δXF
F
F
= 1 + δX ⋅
F
F
= 1 + y δX
F
= 1 + ϵX
F
We define the quantity y δ as ϵ.
Thus,
ϵ=y δ=
d c b
F
+ − −1 ⋅
a a a
F
To understand what this term represents, let us set X = 1 and F to F .
Thus,
ϵ=
F −F
F
We come up with the given interpretation,
ϵ=
change in total moles after complete conversion
total moles fed
We can substitute in the volumetric flow rate equation to get,
v = v (1 + ϵX)
P
P
T
T
The molar flow rate of species j is,
F = F + ν (F X) = F (Θ + ν X)
Here, ν is the stoichiometric coefficient (−1 for A, − for B, + for C, + ) for D and so on.
We thus have the expression for C ,
𝐂𝐣 =
𝐅𝐀𝟎 𝚯𝐣 + 𝛎𝐣 𝐗
𝚯𝐣 + 𝛎𝐣 𝐗
= 𝐂𝐀𝟎
𝐏
𝐓
(𝟏 + 𝛜𝐗)
𝐯𝟎 (𝟏 + 𝛜𝐗) 𝟎
𝐏 𝐓𝟎
𝐏 𝐓𝟎
𝐏𝟎 𝐓
We are now able to expression concentration as a function of conversion. This will help us write the rate law (−r ) in terms
of conversion.
The following chart summarises the idea.
Isothermal Reactor Design: Conversion
These are the general steps we follow when analysing an isothermal reactor.
Derivation of time required for a certain amount of conversion for second order reaction in an Isothermal Batch
Reactor
Let us now derive how much time is required in a Batch Reactor to achieve a specific conversion provided the reaction is
second order with respect to the reactant A.
We have the conversion for BR,
N
dX
= −r V
dt
Note that the volume in a BR is always constant as it is a closed system.
The rate law is second order, thus: r = k C
We substitute,
N
Notice that
dX
= −k C V
dt
= C , we do exactly that,
C
And C = C (1 − X)
dX
= −k C
dt
dX
= −k C (1 − X)
dt
C
Finally,
dX
= −k C dt
(1 − X)
On integrating from 0 → X on the LHS and 0 → t on the RHS, we finally get:
t=
1
k C
X
1−X
The conversion can be written as:
X=
tk C
1+k C t
CSTRs
We know that,
We can write F
V=
F X
(−r )
V=
C v X
(−r )
as C v ,
We can divide this by v to achieve the space time (τ) for a specific conversion:
τ=
V
C X
=
(−r )
v
This is an equation for a single CSTR or the first CSTR connected in series
First Order Reaction in a CSTR
For a first order reaction we have,
τ=
C X
(−k C )
=
C X
X
=
−k C (1 − X) k(1 − X)
The conversion is thus,
X=
C =C
τk
1 + τk
τk
1 + τk
We define a parameter known as the Damköhler parameter for the first order reaction (Da ), as
Da = τk
The equation for conversion becomes,
X=
Da
Da + 1
Second Order Reaction in a CSTR
τ=
C X
(−r )
=
C X
C X
1
X
=
=
⋅
(1 − X)
kC
k(C (1 − X) ) kC
Solving for X, we have:
X=
(1 + 2τkC ) − 1 + 4τkC
2τkC
(We choose the minus sign in the quadratic formula because X ≤ 1, always.
We define the Damköhler parameter (Da ) for the second order reaction as follows,
Da = τkC
The resulting equation for the conversion becomes,
X=
(1 + 2Da ) − 1 + 4τDa
2Da
The Damköhler Parameter
The Damköhler parameter for a reaction is defined as,
Da = −
r V Rate of Reaction at Entrance
Reaction Rate
=
=
F
Entering flow rate of A
Convection Rate
We can see that for first and second order reaction our definition results in the expressions we chose earlier.
Da =
−r V k C V
=
=k τ
F
v C
Da =
k C V
=k C τ
v C
For irreversible reactions,
if Da < 0.1, then X < 0.1
if Da > 10, then X > 0.9
CSTRs in Series
Consider two CSTRs in series as shown below.
For the first CSTR, we have:
C
Solving for C
C
1+τ k
=
we have,
C
=
C
C
=
(1 + τ k )(1 + τ k )
1+τ k
If the reactor volumes are equal, then τ = τ = τ and if operating at same temperature then, k = k = k.
We have,
C
=
C
(1 + τk)
C
=
C
(1 + τk)
For n CSTRs in series,
Substituting for conversion,
C (1 − X) =
X=1−
Derivation for second order reaction in PFR
The differential form of the PFR equation is,
C
(1 + τk)
1
(1 + τk)
F
dX
= −r
dV
dX
−r
V=F
For the second order reaction, we write the rate as −r = k C
dX
=F
k C
V=F
dX
v
X
=
k C (1 − X) C k 1 − X
This gives us,
V
1
X
=
v
C k 1−X
Giving X as,
X=
τkC
1 + τkC
=
Da
1 + Da
Note that this is for liquid phase where we have assumed the volume to be constant, for gas phase we will have to use the
relationship,
v = v (1 + ϵX)
T
T
P
P
That will be the part of the integral.
We can calculate and show that the volume of the reactor will be,
V=
v
kC
2ϵ(1 + ϵ) ln(1 − X) + ϵ X +
(1 + ϵ) X
1−X
Multiple Reactions
There are four cases of multiple reactions.
1. Parallel Reactions
A→B
A→C
2. Series Reactions
A→B→C
3. Complex Reactions
A → 2B + C
A + 2C → 3D
4. Interdependent Reactions
A→C
B→D
Selectivity and Yield
Suppose we have two parallel reactions undergoing,
A + B → D (rate constant: k )
A + B → U (rate constant: k )
We wish to produce D rather than U. Here, the concept of selectivity would help us understand how much the conditions
favour the production of D over U. The selectivity is thus defined as:
S
=
moles of desired product formed
r
=
moles of undesired product formed r
And the yield is defined as,
Y =
moles of product formed
r
=
moles of reactant used
−r
Suppose the reaction of the formation of D is second order in A, and first order in B, while the reaction in the formation of
U is first order in both, we have,
S
=
k C C
k
= C
k C C
k
Thus, the selectivity of D with respect to U is directly proportional to C . Hence, we should have high concentration of A to
favour production of D. We will use PFR instead of a CSTR since the concentration drop in CSTR is much higher for
reactant than in PFR.
Various Contacting Schemes to vary concentrations of A and B
Instantaneous Fractional Yield (𝛟)
For the reaction, A → R
ϕ=
moles of R formed
dC
=
moles of A reacted −dC
Overall Fractional Yield (𝚽)
𝚽=
𝐂𝐑 𝐟
𝐚𝐥𝐥 𝐑 𝐟𝐨𝐫𝐦𝐞𝐝
𝐂𝐑𝐟
=
=
𝐚𝐥𝐥 𝐀 𝐫𝐞𝐚𝐜𝐭𝐞𝐝 𝐂𝐀𝟎 − 𝐂𝐀𝐟 −𝚫𝐂𝐀
Yield for PFR and CSTR
For PFR,
Φ
=−
C
1
−C
ϕdC
For CSTR,
Φ
=ϕ
Enzyme Kinetics
Enzymes are extremely selective catalysts (it can even differentiate between two optical isomers). It is a long chain of
protein.
Enzymes increase the reaction rate; they do NOT change the equilibrium condition. Enzymes change the E , allowing
more molecules to attain the activation energy at the same temperature. It will NOT change 𝐊 𝐂 or 𝚫𝐆. It will however,
change the rate constant.
Enzymes provide an alternate route for the reaction.
We refer to the reactant in enzyme catalysis as the “substrate” (S) and the product formed as “P” the enzyme thus provides
an alternate route between S and P.
E ⋅ S is called the “enzyme-substrate complex”.
The mechanism for an enzyme catalysed reaction is,
Enzyme formation takes place as follows,
E+S→E⋅S
As enzyme formation is a reversible reaction, enzyme-substrate complex will also break down.
E⋅S→E+S
The production formation is generally irreversible
E⋅S+W→P+E
Michaelis-Menten Kinetics
r = k (E ⋅ S)(W)
The rate of formation of E ⋅ S is assumed to be zero as the enzyme does not accumulate,
r
⋅
≈ 0 = k (E)(S) − k (E ⋅ S) − k (E ⋅ S)(W)
(E ⋅ S) =
k (E)(S)
k +k W
The total enzyme concentration is given by,
E = E + (E ⋅ S)
E=
E
k S
1+
k +k W
And now the simplified equation for the reaction rate,
r = k (E ⋅ S)(W) =
(We have defined k W as k
and k
E as V
k WE S
k ES
V S
=
=
k +k W
K +S K +S
+S
k
or the maximum velocity of the reaction and K =
)
k is called the turnover number. It is the number of substrate molecules converted to product in given time by a single
enzyme molecule.
K is the substrate concentration at which the reaction rate is half the maximum rate.
K =S
Monod Equation
The Monod equation describes the growth rate of a bacteria or microorganism in a bioreactor. It is similar to the MichaelisMenten kinetics equation. It is an empirical equation given by,
μ=μ
⋅
[S]
K + [S]
μ − Growth rate of the microorganism
μ
− Maximum growth rate of the microorganism
K − Half velocity constant, i.e., the value of [S] when
= 0.5
[S] − The concentration of the limiting substrate S for growth
Non-Isothermal Reactor Design: Energy Balances
Suppose we conduct the following liquid phase, first order reaction in a PFR,
A→B
The reaction is exothermic, and the reactor is operated adiabatically. As a result, the temperature will increase down the
length of the reactor. Because T varies down the reactor, k will also vary, which was not the case for isothermal PFRs.
Derivation for a non-isothermal PFR
We write the PFR design equation,
dX
= −r
dV
F
And we have,
−r = kC
But know that,
k = k (T ) exp
E 1 1
−
R T
T
Combine,
E 1 1
−
R T
T
−r = k exp
C (1 − X)
And we have,
dX
r
=−
dV
F
=
kC (1 − X)
k
= (1 − X)
F
v
And finally,
dX
=
dV
k exp
E 1 1
−
R T
T
v
(1 − X)
This differential equation cannot be solved directly as it contains three variables (𝐗, 𝐕 and 𝐓).
We thus need another relationship that relates X and T or T and V. This is the energy balance.
If the heat capacities of A and B are equal (C
= C ), we can show that:
T=T +
Now we have everything to solve the differential equation.
The General Energy Balance
−ΔH
C
X
The general energy balance begins with the first law of thermodynamics applied on a closed reactor system. For a closed
system, that is no mass crosses the system boundaries, the energy balance is,
dU̇ = δQ − δW
Here, δQ is the heat energy flowing into the system and δW is the work done by the system.
For an open system, we have the energy balance as shown by the schematic,
dE
= Q̇ − Ẇ + F E − F
dt
E
The unsteady state balance for a system with n species, respective molar flow rate F and with energy per mole, E is,
dE
= Q̇ − Ẇ +
dt
EF
−
EF
The work term is separated into flow work and shaft work. Flow work is the work that is necessary to get the mass into and
out of the system. For example, when shear stresses are absent,
Ẇ = −
F PV
+
F PV
+W
Substituting the expression for work we get:
dE
= Q̇ − Ẇ +
dt
(E + PV )F
−
(E + PV )F
Recall that E is represented as the sum of all energies, the internal energy, kinetic energy, and potential energy.
E =U +
u
+ gz + ⋯
2
In almost all chemical reactor situations, kinetic, potential, and other energy terms are negligible in comparison with the
enthalpy, heat transfer and work terms and hence are omitted. Thus, E = U .
dE
= Q̇ − Ẇ +
dt
(U + PV )F
−
(U + PV )F
−
HF
We recall from basic thermodynamics that,
H = U + PV
Thus,
dE
= Q̇ − Ẇ +
dt
HF
Finally, we can write the general energy balance equation:
𝐝𝐄𝐬𝐲𝐬
= 𝐐̇ − 𝐖̇ 𝐬 +
𝐝𝐭
Here,
H ,F − H ,H ,H
… and F , F , F
H , F − H , H , H … and H , H , H
…
…
𝐧
𝐧
𝐇𝐢𝟎 𝐅𝐢𝟎 −
𝐢 𝟏
𝐇𝐢 𝐅𝐢
𝐢 𝟏
At steady state it can be shown that the energy balance equation becomes,
Q̇ − Ẇ + F
Θ (H − H ) − ΔH (T)F X = 0
Calculating Enthalpies of the Constituents
The enthalpy H is calculated as,
H = H (T ) + ΔH
Here, H is the enthalpy at reference temperature (usually 25 C) and the ΔH
to temperature T,
ΔH
=
is the change in enthalpy after being raised
C dT
C maybe given in the form, α + β T + γ T , then we’d have to integrate and find the answer.
We can show that, provided C is constant over the reference range [T , T],
H − H = C [T − T ]
Thus, the steady state equation can be written as,
Q̇ − Ẇ + F
Θ C (T − T ) − ΔH (T)F X = 0
Calculating Reaction Enthalpies
For the reaction,
b
c
d
A+ B→ C+ D
a
a
a
The reaction enthalpy ΔH (T ) is,
d
c
b
ΔH (T ) = H (T ) + H (T ) − H (T ) − H (T )
a
a
a
Similarly, the change in heat capacity, ΔC for the reaction is given by,
d
ΔC = C
a
c
+ C
a
b
− C
a
−C
Combining the equations we have,
ΔH (T) = ΔH (T ) + ΔC (T − T )
Adiabatic Equilibrium Conversion
The equilibrium conversion is the maximum possible conversion in a reversible reaction. For an exothermic reaction, the
equilibrium conversion increases as temperature decreases, for an endothermic reaction it is the reverse.
Equilibrium curve for exothermic reaction
Equilibrium curve for endothermic reaction
Optimum Temperature Progression
The Residence Time Distribution (RTD)
So far, our analysis of reactors has largely dealt with ideal reactors. However, in the real world, no reactor is perfectly
mixed or ideal.
For example, look at this diagram of a zoomed in cross-section of a packed bed reactor:
While we have assumed the packed bed reactor to be a continuous solid, this is not the case in real life. Regions of the solid
may provide low resistance to the incoming fluid, like Path 1, and some may provide more resistance, as in Path 2.
Consequently, we see that there will be a variation in time which molecules will be in contact with the catalyst. Some
molecules will remain in contact for more time while some will remain for less.
In another example, we consider a CSTR whose entry and exit pathways are close to each other. In this case we encounter
“dead zones”. In these regions there is no exchange of material with well-mixed regions and as such no reaction occurs
there. This phenomenon is called bypassing.
The Residence Time
The residence time is defined for an atom or molecule. The residence time is the amount of time that an atom or molecule
has remained inside the reactor.
In an ideal situation, all the molecules leaving a PFR at time t = t have been in that reactor for an identical amount of time.
No variation exists. In the same way, in a batch reactor, we assume that all the atoms or molecules have remained inside for
an identical amount of time, between t = 0 and t = t i.e., when we have achieved the required conversion and removed the
reactants and products.
Measurement of the RTD: The Tracer
To measure the RTD of a reactor, we inject a tracer. This tracer is inert, easily detectable (usually is coloured, or
radioactive), usually a noble gas, and has physical properties like those of the reacting mixture and be completely soluble in
the mixture. It also should not adsorb on the walls or other surfaces of the reactor.
The Pulse Input Experiment
In a pulse input, an amount of tracer N is suddenly injected in one shot into the feed stream entering the reactor in a very
short amount of time as humanly possible. The outlet concentration of the tracer is then measured as a function of time.
The effluent concentration curve versus time is referred to as the C-curve.
First, we analyse the tracer pulse for a single output system in which only flow, and no dispersion carries the tracer material
across the system. We choose a time Δt, sufficiently small such that the concentration of the tracer, C(t), exiting between
time t and t + Δt is constant. The amount of tracer material leaving the reactor in t and t + Δt is:
ΔN = C(t)vΔt
Here v is the volumetric outflow rate.
ΔN is the amount of material exiting the reactor that has spent an amount of time between t and t + Δt in the reactor. If we
divide by the total amount material that was injected into the reactor, we obtain,
ΔN vC(t)
=
Δt = E(t)Δt
N
N
We define the E-function as
()
.
This quantity E(t) is called the residence time distribution function. The quantity E(t)dt is the fraction of molecules that
have spent an amount of time between t and t + dt inside the reactor.
If N is not known directly, it can be determined from the C curve by integrating it to infinity. We have,
dN = vC(t)dt
And then integrating, we obtain,
N =
vC(t)dt
The volumetric flow rate is usually constant, and we have, E(t) =
E(t) =
()
, substituting the value of N we get:
C(t)
∫ C(t)dt
[Fraction of Material Having Residence Time Between t and t ] =
Also, from the definition we have the following result:
E(t)dt
E(t)dt = 1
The F-function
We define a function F(t) for a corresponding E(t) that tells us the fraction of molecules or atoms that have spent a
residence time between t = 0 and t = t.
F(t) =
E(t)dt
The fraction of molecules that have spent residence time more than t thus becomes 1 − F(t).
Note that the definition also means,
E(t) =
dF
dt
Step Tracer Experiment
In a step tracer, instead of providing tracer input at one instant of time, we provide it continuously. Stated symbolically,
C (t) =
0
t<0
C , constant t ≥ 0
Because the inlet concentration is constant, we can write it as:
(t) = C
C
(t)
C
C
F(t) =
E(t) =
E(t )dt′
=
E(t )dt
C
(t)
C
dF d C (t)
=
dt dt C
The positive step is usually easier to carry out experimentally than the pulse test and it has the additional advantage that the
total amount of tracer in the feed over the period of the test does not have to be known as it does in the pulse test. The
drawback is that one must maintain constant tracer concentration in the feed. Another problem is that the technique requires
differentiation which can lead to large number of errors. And finally, the tracer amount required for this test is much larger
than the one required for the pulse test.
Mean Residence Time
We define the mean residence time, 𝐭 𝐦 as:
t
=
∫ tE(t)dt
∫ E(t)dt
=
E(t)dt
It can be mathematically shown that the mean residence time for a closed system (i.e., no dispersion across boundaries) the
mean residence time is equal to the space time (we will not do this proof):
𝐭𝐦 = 𝛕 =
[reactor volume]
𝐕
=
𝐯 [volumetric flow rate]
Thus, the volume of the reactor can be determined:
V = vt
Standard Deviation of the RTD
The standard deviation which is the square root of the variance can also be determined.
(t − t ) E(t)dt
σ =
S. D. =
(t − t ) E(t)dt
σ =
Skewness of the RTD
The skewness (s ) is defined as,
s =
1
(t − t ) E(t)dt
σ
The Normalized RTD Function, 𝐄(𝚯)
We define a parameter Θ as,
Θ=
t
t
=
t
τ
The quantity, Θ represents the number of reactor volumes of fluid that have flowed through the reactor in time t. The
dimensionless RTD function, E(Θ) is then defined by,
t
E(Θ) = E(t) = τE(t)
τ
It is easy to show that,
E(Θ)dΘ = 1
RTD for Ideal PFR and Batch Reactor
For the PFR and Batch Reactor, as the reactants simply go in and out, and all the tracer comes out at one instant. The E(t) at
τ is infinity.
The RTD function is thus the Dirac delta function shifted by τ.
E(t) = δ(t − τ)
Recall that the Dirac delta function is,
δ(x) =
0 when x ≠ 0
∞ when x = 0
Some properties of the Dirac delta,
δ(x)dx = 1
g(x)δ(x − τ)dx = g(τ)
The second property helps us calculate the mean residence time, have g(x) = t.
tE(t)dt =
tδ(t − τ)dt = τ
The standard deviation is 0 as can be shown,
(t − τ) δ(t − τ)dt = (τ − τ) = 0
σ =
All of this can be confirmed by intuition.
RTD for single CSTR
For PFR and BR, since the reactants go in and out, the RTD curve was very simple. The CSTR is a bit different, as here, the
reactants are being constantly mixed, and the tracer will also be completely mixed into the reactor. Thus, it will not all come
out at one instant. We know that the concentration inside a CSTR at any point is C, thus the amount of tracer coming out of
the CSTR will be Cv where v is the volumetric flow rate. This will induce a change in the tracer concentration within the
reactor as
, and therefore the change in the number of moles left inside the CSTR will be V ⋅
A material balance for the inert tracer that has been injected as a pulse at t = 0 for t > 0 gives:
0[
]
− vC[
]
=V
dC
dt
[
]
Solving this simple differential equation we get,
C(t) = C e
The quantity
= τ or the space time, thus:
C(t) = C e
We can calculate E(t),
E(t) =
C e
∫ C e
We have, E(Θ) = τE(t) = e
=e
=
e
τ
.
The cumulative distribution curve F(t) is,
F(t) =
e
= 1−e
τ
F(Θ) = 1 − e