Idealized enzyme reactor performance Video Lecture - Biotechnology Engineering (BT)

Today we shall discuss the performance of
enzyme catalyzed reactors, the reactors in
which enzyme catalyzed reactions are supposed
to take place.You will recall earlier we had
discussed some of the salient features of
different reactor types ranging from batch
reactors, batch stirred tank reactors to continuous
stirred tank reactor or plug flow reactors
and fluidized bed reactors. In fact some of
the unusual reactor configurations which have
been proposed to catalyze to be used for carrying
out enzyme catalyzed reaction like hollow
fiber reactors also arbitrarily fall into
one of those categories of plug flow or CSTR
which were discussed. We had also seen earlier
the factors that affect the choice of different
reactor types and to recall some of the factors
that we had talked about the most important
ones are listed here like form of the enzyme.
That is soluble, fibrous or particulate. The
nature of the substrate whether it is soluble,
particulate or colloidal, operational requirements;
the control of pH in case an acid is produced
or consumed in the reaction, reaction kinetics,
carrier loading capacity, catalytic surface
to reactor volume ratio, mass transfer characteristics,
ease of catalyst replacement and regeneration,
reactor cost and ease of fabrication. All
these factors that I have listed here can
be considered mainly under two categories.
One category contains the factor which becomes
fixed once an immobilized enzyme is prepared.
That means the factors that are dictated by
the immobilization enzyme preparation itself.
For example form of the enzyme; the nature
of the substrate is also fixed. Once the enzymatic
reaction is defined the nature of substrate
is fixed. The operational requirement is also
fixed depending upon the nature of the enzyme
catalyzed reaction whether an acid is produced
or consumed and pH control is a requirement.
Similarly the catalytic surface to volume
ratio is also fixed once an immobilized enzyme
preparation is available. The carrier loading
capacity also depends on the bulk density
of the immobilized enzyme preparation and
things like that. Ease of catalyst replacement
and reactor cost and ease of fabrication are
the factors which are dependent on the type
of enzyme reactor. But the two important class
of factors which I wanted to point out here
is one is reaction kinetics and mass transfer
characteristics. They are the ones which very
heavily dictate the choice of the reactor;
whether to go for a plug flow reactor or to
go for a batch reactor or to go for a continuous
stirred tank reactor are very heavily dictated
by mass transfer requirements and reaction
kinetics. I would once again point out that
we have knowingly ignored heat transfer requirements
assuming that for most of the enzyme catalyzed
reaction the rate of reaction is very, very
small and so heat transfer during the reaction
is not a major critical factor. In certain
cases if it is so, the heat transfer characteristics
also must be taken into account. Last time
I was taking we had also considered certain
idealized reactor design parameter. That means
for design of a particular type of reactor
what kind of design parameters are important
and you will recall that we had also hypothesized
certain conditions or assumptions which define
an idealized enzyme reactor system and the
assumptions were that isothermal operation,
?H is very small, uniform enzyme distribution,
plug flow behavior or completely back mix
behavior, that means you have one of the extreme
fluid dynamics in the reactor system. Either
it is a plug flow motion or it is a completely
back mix and nothing in between. That is what
we have assumed and there are no mass transfer
limitations and no significant partitioning
of substrate between bulk and the carrier
phase.
These are the basic assumptions which define
our ideal enzyme reactor and the reactor design
parameters. That means the parameter which
we should define initially before we go on
to design any particular reactor was “tau”,
the space time in the reactor; initial substrate
concentration, enzyme loading, E0. In the
case of continuous operation, the information
about the deactivation rate constant is also
required, kd. Then the temperature of operation,
pH control and these are the reaction conditions.
Other design parameters are fractional conversion
and productivity and reactor capacity.
That is the maximum reaction capacity assuming
that the enzyme reaction follows a maximum
reaction velocity, that is zero order profile
and the productivity of that kind of the reactor
will be your reaction capacity. Ultimately
the performance will be dictated very heavily
by enzyme kinetics. If we consider ….. enzyme
reactor we can safely say that Michaelis Menten
kinetics provides us a basis for carrying
out a universally accepted kinetics for enzyme
catalyzed reactions and so therefore we will
assume in most cases v is equal to
v = k2E0S/Km+S
Wherever we are talking of immobilized enzyme
we must also clearly understand that these
two terms k2 and Km are denoted by prime indicating
only that they are immobilized enzyme parameters
and not the soluble parameters.
The first and probably the most simplest reactors
is the batch reactors. The batch reactor can
be used both for soluble enzyme as well as
for immobilized enzyme.You can carry out a
batch reactor in a stirred mode or in a packed
bed mode with using total recycle and on the
right hand side is the characteristic profile
of the reaction species that is the substrate
and product with reference to time. The substrate
concentration drops and ultimately depending
upon the extent of conversion which we require,
the product concentration increases. A typical
profile; it can change from reaction to reaction.
If the reaction is very strongly inhibited
by products or strongly inhibited by substrates
the profile might undergo a change. One of
the probably very general features of analyzing
the reactor performance as you may be aware
is material balance across the reactor. You
take material balance for one of the reaction
species say substrate and make a material
balance; simplify it. Do mathematical steps
and you will arrive at the reactor performance
equations which will give you the performance
of reactor. When I say performance of reactor
I mean the extent of conversion as a function
of space time or in a batch reactor with the
reaction time. In the case of a batch reactor
the reactor performance can be just simplified
integrated form of the reaction equation and
that is
-ds/dt = k2.E0
Km/S + 1
Typical Michaelis Menten equation. Even when
we write integrated form of equation we can
assume that this is also valid if you make
a material balance across the batch reactor
because the classical material balance equation
requires that input minus output minus conversion
should be equal to accumulation. Here in the
case of a batch reactor the accumulation is
volume multiplied by rate of change of substrate
concentration that is –v.ds/dt. Input and
output here are taken as zero because there
is no continuous input and output. Only initially
we feed the reactor and at the end of the
reaction time the output is taken out. There
is no continuous output or input in the system
and the conversion is given by the rate of
reaction. So v into rate of reaction is the
total accumulation and the conversion factor
and by canceling the volume term from both
the sides you can arrive at the integrated
form of the equation and this equation if
you integrate, it will lead to the reactor
performance equation.
If you just simplify it, you will get
Km ln S0/S + S0-S = k2.E0.t
Again I like to repeat that in case if we
are using a batch reactor for immobilized
enzymes we should replace Km and k2 by prime
terms. This is equation 1 and this is equation
2. If you want to describe the equations in
terms of reactor performance then X is equal
to
S0-S
X = = 1- S/S0; S/S0 = 1-X
S0
So if you substitute all these parameters
you get in terms of reactor performance equation
for a batch reactor in the form of fractional
conversion that is
XS0 + K’mln 1/1-X = k’2E0t
The reaction performance equation for a batch
reactor if we want in the terms of product
concentration, product concentration is nothing
else but
P= S0-S
assuming that its a unimolecular reaction.
If in case you are handling with a different
kind of a reaction scheme let us say one mole
of substrates goes into two moles of products
then accordingly one has to make a change
in the definition of X as well as P. This
will be equal to
P= S0-S = XS0
Therefore you can also write in terms of product
concentration
KmlnS0/S0+P = k’2E0t
If you
look at these equations you can use the reactor
performance data of a batch reactor. That
means you take a single batch performance
data; take the maximum conversion at different
times of reaction. You can write down the
expression
ln S0/S 1 k’2E0t
+ =
S0-S Km K’m (S0-S)
From the equation two if you just simplify
you get a straight line equation which can
give the values of Km and vm. If you plot
let us say ln(S0-S)/S0-S and t/ S0-S, you
get the slope of k’2E0/ K’m and
with the intercept as -1/ K’m
Sorry. X-axis is lnS0/S. This provides you
one of the simple methods. When we immobilize
enzymes if we monitor the kinetics parameters
Km and vm, the simple way to measure the kinetic
constants for an immobilized enzyme reaction
will be to carry out the reaction in a batch
stirred reactor. Monitor the substrate concentration
or fraction of conversion with reference to
time and determine the parameters by a single
reactor performance data. That provides us
a simple tool, batch reactor for small scale
laboratory experiments to determine the kinetic
parameters both for soluble enzyme as well
as for immobilized enzyme.
Here I must again put a caution. When we used
the Michaelis Menten equation you will recall
that we had made certain assumptions and the
assumptions were that we are talking of the
initial reaction rate. That means at a time
equal to zero. Whereas when we integrate the
equation from zero to time t let us say to
a conversion factor of 90% or 95% or 99%,
the reaction rate is not likely to remain
the same as that of the initial time. So here
we are violating some of the basic assumptions
of the Michaelis Menten equation. But still
it gives you a very approximate situation
for designing a reactor or making some design
calculations to work out reactor size or the
capacity of the reactor, accepting that and
even compare between soluble enzyme and immobilized
enzyme but otherwise under idealistic conditions,
the assumptions in the Michaelis Menten equation
that of the initial reaction rate, absence
of product division is not met here. So when
we apply the integrated Michaelis Menten equation
for looking at the performance of an immobilized
enzyme reactor in the batch mode we must keep
in mind that it only gives you an approximate
situation just to tell the sizing of the vessels
required for a particular enzyme catalyzed
reaction and to compare the soluble enzyme
with that of immobilized enzyme performance.
Even while looking at the kinetics of the
soluble enzyme we always look at the situations
when it comes to extreme substrate conditions.
That means either when the substrate concentration
is much, much smaller than Km or when it is
much higher than Km. That means it ranges
from first order to zero order regime. How
the reactor performance will change? For example
when substrate concentration is much, much
smaller than K’m, reactor performance becomes
-K’m ln(1-X) = k’2E0t
When S0 is much, much greater than K’m that
is zero order regime.
XS0 = k’2E0t
If suppose you approximate an enzyme reactor
to follow either a zero order or a first order
regime, knowing the values of Km and k2 the
error involved in that will be determined
by the ratio of Km/S0. If the ratio of Km/S0
is very, very small you will hardly incur
any error if you assume it to be a first order.
When we go on to consider along with the reaction
kinetics we consider mass transfer and all
more complications we will tend to approximate
our reactor performance in either zero order
or first order regime rather than making it
a Michaelis Menten equation which might make
the solution of the expressions more difficult
or it can be done by any numerical methods.
But analytical solutions may be difficult.
We make an assumption depending upon the range
in which our substrate concentration exists
in relation to Km value. The error involved
will increase if you increase the ratio of
Km to S0 and you assume a first order because
then both will go towards the zero order kinetics.
If the Km to S0 ratio increases and you assume
a first order kinetics error will keep on
increasing. Km is much, much larger than the
S0 in the first order. So the value of Km/S0
will be larger. Km/S0 will be large and under
that condition you must assume a first order
kinetics and if the value of Km/S0 decreases
the error will continue to increase. In fact
one can make a hypothetical computation and
then we take up an assignment to design an
enzyme reactor or for that matter any process
equipment. A 10% error in calculation is usually
taken for granted. If you are going to measure
let us say the value of Km or vm, you need
an accurate value. But when you want to go
for design, let us say a reaction capacity;
what is the volume of the reactor required
to carry out this conversion? You are not
going to be very specific that this is the
volume required. You will add on to that about
10% open added space or for various purposes.
So if let us say the error involved is less
than 10%, I think there is no harm in making
an assumption of zero order or first order.
If the error involved is much larger then
our assumptions will be slightly hazardous.
The next simple form of reactor system which
is commonly employed is a continuous stirred
tank reactor. In the continuous stirred tank
reactor one is a continuous flow reactor.
That means the feed rate of the substrate
is same as the output rate of the product
stream. So if the feed is added at the flow
rate of “F” let us say litres or moles
per unit time, what ever unit you follow,
the volume remains the constant. Substrate
concentration S0, changes at the exit. But
one of the characteristic features is that
all the parameters at the exit stream are
same as in the reactor itself. That means
the reaction will take place under conditions
of S, P or X which exist at the exit stream.
Therefore if you just write a material balance
for such a reactor system you can write accumulation
as
V.ds/dt = FS0 –FS – v.V
FS0 is the input; S is the substrate concentration
in the outlet and v is the reaction velocity
that is the conversion due to the reaction
and V is the reactor volume. Consider that
after a certain period of time the reactor
comes under steady state situation. When we
say steady state situation it means that the
concentration of any species does not change
with time in the outlet stream. Under that
condition the accumulation term will tend
to be zero because whatever material is coming
in, either it is getting converted to do reaction
or it is going out of the reactor. There is
no accumulation of any species in the system
and therefore one can write
k’2E0S
v = D(S0-S) =
K’m + S
D is
the term here what we understand as dilution
rate or F/V or in other words V/F is the resistance
time, inverse of space time. So this D is
the term and you can also write
k’2E0S S0 - S
=
K’m + S ?
That
is
your space time that was defined in the earlier
expressions. Solving these equations you can
also get
k’2E0 ? = S0X + K’m (X/1-X)
That gives you a reactor performance for a
CSTR and from the substrate concentration
if you make a substitution for X, you arrive
at detector performance for CSTR. Similarly
here also if you make assumptions of zero
order or first order situation when K’m
is much, much smaller than S0, that is a zero
order system
? = XS0/k’2S0
For first order system where K’m is much,
much greater than S0, you have
K’m(X/1-X) = k’2E0?.
Second expression is ? = XS0/k’2E0. So you
get the two expressions for CSTR the general
reactor performance expression and also under
conditions of zero order and first order scenario.
The expressions are very simple and if you
have the data available for any reactor performance
for S0, fraction of conversion required, you
can determine the volume of the reactor required
or the feed rate required depending on the
tau and one can make computation to define
the reactor performance under given conditions.
The other idealized reactor or which is what
we understand as the plug flow is slightly
typical in the sense that unlike in the case
of a CSTR, the concentration of the substrate
or the product vary along the length of the
reactor. So making a material balance across
the reactor you cannot choose any concentration
term because the concentration term for substrate
changes with the length of the reactor and
the concentration of the substrate or product
is not constant throughout the reactor. It
varies; that means if you consider along the
length of the reactor, kind of a batch reactor
with different reaction times but to arrive
at the reactor performance what we can do
is we can make a material balance across a
differential element of the plug flow reactor
and if we consider differential element of
volume d v you can make your material balance
say under same conditions like the feed flow
rate as “F”, the conversion changes from
X to X +?X at the inlet X is equal to zero
at the outlet X has some value.
If you consider here input to the system is
F.S and this input I am writing at the entry
of the differential element and output, not
at the entry of the reactor, is equal to F(S+dS)
and the disappearance of the substrate by
the reaction will be the continuation of the
reaction to the disappearance of the substrate
is
–v.(dV) Under these conditions one can write
the material balance
under steady state assuming that the accumulation
is zero.
F.S = F(S+dS) + (-v).dV
One can simplify it to get
F.dS = v.dV
Here for the fraction of conversion
X = S0-S/S0; dS = -S0dX
F.S0.dX = -v.dV
You can integrate this and
the integrated form will give you the zero
to v dv /FS0 which will be equal to zero to
X dX/-v or V/FS0 because feed rate is constant,
substrate concentration is constant and so
the definite integral will come to V/FS0 and
that is ?/S0 which is equal to zero to X dX/-v.
If you look at the performance of a plug flow
reactor you ultimately arrive at or transfer
the S0 to here so tau is equal to S0.dX/-v.
That is the reaction rate.
So the performance will be almost identical
to that of
a batch reactor excepting with the difference
that the reaction time here is replaced by
the space time and the reactor performance
therefore on the same lines you can write
as
S0X – K’m ln (1-X) = k’2E0 e ?
Epsilon, while in the case of a stirred bed
reactor, is taken to be one. That means the
volume of the reactor occupied by the enzyme
preparation is negligible compared to the
total reactor volume whereas in the case of
a packed bed a significant proportion of the
reactor is occupied by the enzyme preparation.
I was just trying to define the term what
I introduced here as epsilon. In the case
of a batch stirred reactor we have not considered
epsilon because the quantity of enzyme or
the volume of enzyme in the reactor is negligible
compared to the total volume of the reactor.
A very small fraction of the total volume
is occupied by the enzyme preparation. On
the other hand in the case of a packed bed
reactor a significant fraction of the volume
of the reactor is occupied by the packed bed
and the volume occupied by the fluid or the
feed substrate is only the wide volume of
the reactor and therefore the volume occupied
by the feed in the reactor is considered as
the actual reactor volume. This tau term has
been defined earlier as V/F; in the case of
a packed bed reactor we define tau as eV/F.
That is for CSTR or for a stirred batch reactor
the magnitude of epsilon is one. So this gives
you the reactor performance for a plug flow
reactor following Michaelis Menten kinetics.
The same analysis as we did earlier if you
consider for zero order kinetics the tau is
equal to
? = XS0/k’2E0e
For first order, tau is equal to
? = -K’m ln (1-X) / k’2E0e
A very important feature which I like to recall
is the relative performance equations for
zero order regime for both CSTR as well as
for PFR, which are identical. If you look
at the equation number two here for the CSTR
and the equation for the zero order regime
here in the case of the PFR, they are identical.
That means the fluid dynamics in the reactor
whether it is a back mix system or its a plug
flow system does not influence the reactor
performance as long as the reaction is in
the zero order regime or the concentration
of the substrate is much, much higher than
the Km value and you must appreciate that
if it is in our limits we will always like
to use a very high substrate concentration.
If the system doesn’t undergo substrate
inhibition that is if it undergoes inhibition,
we have no choice other than to using a lower
substrate concentration. There are two things
one will always like to have a high substrate
concentration for a reaction; for commercial
purposes also we like to have a very high
fraction of conversion and one would like
to carry out the reaction almost to conversion
levels of ninety five percent plus so as to
achieve a complete conversion of substrate
as much as possible so that the recovery and
the purification stages are simplified or
the product stream can be used as it is for
end applications. So the only difference lies
in the case of reactor performance of plug
flow reactor. Consider a
first order regime where K’m is much, much
greater than S0. In the case of CSTR, you
have the reactor performance as
K’m(X/1-X) = k’2E0?
In the case of PFR, the expression changes
to
K’m ln (1-X) = k’2E0 e ?
Our interest will always be to have a high
fractional conversion. That means most of
the substrates which is being fed gets converted
into product. If we want to compare the performance
of both CSTR and PFR as I mentioned in the
case of zero order regime there is no change
and the performance remains identical. That
means the quantity of the enzyme required
will be identical in both cases for carrying
out the same conversion and at the same enzyme
loading. But if you consider the first order
regime the performance are quite different
and if you compare the relative performance,
relative quantity of the enzyme required that
is ECSTR, the quantity of enzyme required
in CSTR and PFR relative quantities or we
can define this term as Erel will be equal
to
ECSTR/ EPFR = Erel = - X e/(1-X) ln (1-X)
The relative quantity of enzyme required in
the case of CSTR and PFR under first order
regime can be given by this expression. You
notice in this expression that higher the
required conversion level, higher the relative
amount of the enzyme required in the CSTR.
You can just hypothetically put some values
of let us say a safe value for epsilon for
a typical packed bed reactor varies from between
0.4 to0.5. You take a simple value of 0.5
and if you make calculation for X=0.90, the
Erel required will be equal to 4.5. That means
the enzyme required for the same fractional
conversion that is 90% in the case of a CSTR
will be 4.5 times more than that required
in the PFR for the same conversion. The same
enzyme, the same reaction, the same conversion
factor but the quantity of enzyme required
in CSTR will be much, much higher compared
to PFR. If you change the fractional conversion
to, let us say, 0.99, the Erel value shoots
up to 25.
The inference I want to arrive at is that
higher the fractional conversion required
for a particular reaction under first order
regime, the relative quantity of the enzyme
required in the CSTR will drastically increase.
That means even for certain reasons if you
want to use CSTR pH control is a requirement
and we are obliged to use CSTR, then it is
always advisable only to use if the fraction
of conversion required is not very high. In
case if the fraction of conversion required
is very high if you are looking for a 99%
conversion, it will be desirable to use a
plug flow reactor with a partial recycle.
You can have facility for pH control with
some inconvenience; may be some more additional
equipments has to be added but still that
will be better because the quantity of the
enzyme required will be much less and that
gives us a significant advantage for using
immobilized enzyme in PFR and that is one
of the reasons why you will notice in the
literature that almost about 70- 80% of the
total reports which appear on the use of the
immobilized enzymes are in a plug flow reactors
and the advantage lies in the quantity of
the enzyme required and this advantage will
keep on reducing. That is the relative quantity
will keep on reducing if the shift is from
first order to zero order ultimately it will
reach to same quantity. If we take the value
of Km/S0 from a very high value to a very
low value, the relative quantity of the enzyme
will also keep on reducing and it will become
equal to the relative enzyme value, Erel and
will become one. Then the value of S0 is much,
much higher than Km.
A typical profile for relative quantity required
is shown here. As you notice here what ever
I mentioned, it signifies the same thing.
There are two things to be looked at. One
is the relative quantity of substrate to Km.
Here instead of Km/S0 I plotted S0/Km. The
top profile, the first profile indicates a
very low value of S0/Km which means a first
order regime. That means as the fractional
conversion increases, the relative quantity
required also increases and towards the final
stages of the fractional conversion, the quantity
very significantly goes up. The increase in
the quantity of the enzyme required is not
very large in the low fractional conversion
levels whereas as soon as you cross 0.90%
conversion, the quantity required becomes
significantly high.
On the other hand you also see when you come
to S0/Km as hundred approaching towards zero
order regime the quantities remain almost
equal all through, till you arrive at 90 plus
or more than 95% fractional conversion.
One can have such a profile and
when one wants to analyze the reactor performance
for any given immobilized enzyme system, one
must really characterize a system with reference
to such kind of profiles so that one can get
a feel of the substrate concentration required
and the fractional conversion required, one
can choose where to bank upon. For example
in the earlier phases let us say up to 90%,
there is no major gain in using PFR. But if
you want to use a fraction of conversion much
higher than 90% then the PFR becomes more
and more matter of choice. Then the immobilization
will be identical. Your choice should depend
upon the operational convenience. For example
if the pH control is required CSTR is preferable
and then the nature of the substrate, the
nature of the enzyme and the form of the enzyme.
If suppose your substrate is colloidal, we
will prefer to use CSTR rather than the PFR;
it will choke. The pressure drop will continuously
keep on increasing. If suppose the enzyme
particles are in the form of fibers or spherical
beads, it is easier to pack them rather than
putting in the CSTR where there might be abrasion.
So the form of substrate and form of enzyme
will dictate the choice. The kinetics will
not be dictating. So far we have not considered
the inhibition patterns. We have not considered
the behaviour of the reactor performance if
the enzyme undergoes inhibition kinetics.
That means either a substrate inhibition or
a product inhibition or a third inhibitor
an external inhibitor available in that. We
have also not considered the deactivation
of the enzyme on the reactor performance.
All these parameters particularly the inhibitors,
rather their inhibition pattern, their deactivation
pattern, the mass transfer requirements, the
mass transfer limitations and the non-ideal
flow behavior; these are the four things.
So far we have been talking assuming that
the enzyme reactors are ideal in performance
and they are following Michaelis Menten type
of kinetics. At best the extremes of those
cases under zero and first order but we have
not considered the inhibition pattern, we
have not considered the deactivation kinetics,
mass transfer limitations
and non-ideal flow behavior. That is we are
considering only either a PFR or CSTR which
is an ideal flow dynamics in the reactor.
In many cases practical reactors may not be
an ideal CSTR or ideal PFR. They might be
some kind of deviation from the ideality because
of the packing characteristics or some channeling
taking place or in laboratory you can very
easily maintain a good plug flow reactor.
But in a large system it is always difficult
to maintain and it is always desirable to
identify or characterize your reactor with
reference to flow behavior and then make a
correction factor when the design part comes
into. So all these issues we will take individually
rather than compounding all of them because
not that all the issues will be applied in
a particular reactor system we will have to
identify the issues which are important. For
example we identify let us say the mass transfer
limitations are the most acute. The system
is mass transfer limited. There is point to
consider reaction kinetics; you design your
reactor system based on mass transfer diffusional
requirements. Similarly if you consider that
the flow behavior is drastically non-ideal
there is a full channeling; then corrections
are to be applied to the design expression
to get the reactor performance. I think we
will stop here and continue to discuss the
behavior of enzyme reactors.