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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.