Question 13.4: Extended Culture (Fed Batch) Operation of a Well Agitated Bi...

Equations (I), (J), and (K) of Illustration 13.3 may be used to determine both the amount of microorganism present at the time when the conversion of the limiting substrate reaches 98% and the time necessary to achieve this conversion. At 98% conversion of the limiting substrate,

S _{98}=(1-0.98) S _0=0.02(35)=0.70  g

and the total mass of the microorganism is

From Table I13.1

for K_{S} = 2.78 g/L, the corresponding value of K_S /\left\{ S _0+\left[ X _0 /\left(V_R Y_{ X / S }\right)\right]\right\} is 0.0722. Recall that equation (I)  t_{\text {batch }}=\frac{1}{\mu_{\max }}\left\{\left[\frac{K_S}{s_0+\left[ X _0 /\left(V_R Y_{ X / S }\right)\right]}\right] \times\right. \left.\ln \left(\frac{ X s_0}{ X _0 s}\right)+\ln \left(\frac{ X }{ X _0}\right)\right\} of Illustration 13.3 is

Substitution of the values of S and X at 98% conversion of the limiting substrate into the preceding equation gives

Thus, the operator begins feeding the bioreactor with a stream containing 50.0 g/L of the limiting substrate 21.47 h after the growth medium is inoculated with the microorganism.

Now we need to consider the equations that describe the operation of the bioreactor once the feed of fresh substrate to the reactor is initiated. For subsequent times, the volume of the suspension of the microorganism will be given by a relation of the form

V_R=V_{R 0}+\int_{t_{98}}^t V d t               (A)

where V_{R} \text{ and  } V_{R0} are the volumes of the suspension at any time (t) and initially, respectively. The parameter t_{98} is the time that it takes the contents of the bioreactor to achieve 98% conversion of the limiting substrate while operating in a strictly batch mode. The parameter V is the volumetric flow rate of the stream fed to the bioreactor to make up for the substrate that has been consumed by the bioconversion process. This flow rate will be controlled to match the rate of consumption of fresh substrate in the bioreactor.

For pedagogical purposes in Illustrations 13.3 and 13.4, we arbitrarily selected 98% conversion of the limiting substrate as the level of conversion at which the change from pure batch to extended culture operation is made. However, in industrial practice determination of the level of conversion at which the transition should be made is a fairly complex optimization problem that is beyond the scope of this book. Additional analyses of fed batch operation may be found in books by Nielsen and Villadsen (14) and Lim and Shin (15).

We next wish to perform a material balance over the bioreactor for the microorganism assuming that the feed contains substrate but not biomass (i.e., it is sterile). If biomass neither enters nor leaves the bioreactor and if we assume that the rate of cell death is negligible compared to the rate of cell growth, we find that

\frac{d\left[x V_R\right]}{d t}=Y_{ X / S } \mu x V_R               (B)

or

x \frac{d V_R}{d t}+V_R \frac{d x}{d t}=Y_{ X / S } \mu x V_R             (C)

where x is the mass concentration of the microorganism in the bioreactor. Combination of equations (A) and (C) gives

x V+V_R \frac{d x}{d t}=Y_{ X / S } \mu x V_R               (D)

Rearrangement yields

\frac{d x}{d t}=x\left(Y_{ X / S } \mu-\frac{V}{V_R}\right)                  (E)

The ratio of the volumetric flow rate to the volume occupied by the suspension is defined by microbiologists as the dilution rate (D).

D=\frac{V}{V_R}=\frac{V}{V_{R 0}+\int_{t_{98}}^t V d t}             (F)

In fed batch systems when the volumetric flow rate of the feed stream varies in a manner that maintains the concentration of the limiting substrate constant, the dilution rate must increase as time elapses because the volume occupied by the aqueous solution increases as more feed enters. Combination of equations (E) and (F) gives

\frac{d x}{d t}=x\left(Y_{ X / S } \mu-D\right)            (G)

The rate at which the substrate is consumed by the bioconversion process will be assumed to be described by a Monod relation:

-r_s=\left(\frac{\mu_{\max } s}{K_S+s}\right) x=\left(\frac{\mu_{\max } s_{98}}{K_S+s_{98}}\right) x           (H)

where we have recognized that the substrate concentration is a constant for the fed-batch portion of the trial and that the reaction remains autocatalytic. Moreover,

r_{ x }=Y_{ X / S }\left(-r_s\right)=Y_{ X / S }\left(\frac{\mu_{\max } s_{98}}{K_S+s_{98}}\right) x                   (I)

Because we know the concentrations of substrate and biomass present in the bioreactor at the moment the fedbatch operation begins (0.70 and 26.82 g/L, respectively), we can employ equation (H) to calculate the rate at which
substrate is being consumed at this time.

-r_S=\left(\frac{\mu_{\max } s_{98}}{K_S+s_{98}}\right) X=\left[\frac{3.16(0.70)}{2.78+0.70}\right]= 17.05  g/(L⋅day)               (J)

This consumption rate determines the initial rate at which substrate must be fed to the bioreactor if the concentration of substrate in the reactor is to remain constant at 0.70 g/L.

A material balance on the substrate for times greater than 21.47 h leads to the equation

\frac{d\left[s_{98} V_R\right]}{d t}=Vs_{\text {feed }}-\left(\frac{\mu_{\max } s_{98}}{K_S+s_{98}}\right) x V_R      (K)

or

s_{98} \frac{d\left[V_R\right]}{d t}+V_R \frac{d s_{98}}{d t}=s_{98} \nu+V_R \frac{d s_{98}}{d t}=V s_{\text {feed }}-\left(\frac{\mu_{\max } s_{98}}{K_S+s_{98}}\right) x V_R                (L)

where s_{98} refers to the constant concentration of substrate in the well-agitated growth medium. However, the mode of operation has been chosen to maintain the concentration of substrate in the aqueous phase constant (ds_{98}/dt = 0) and the time derivative of the reactor volume is equal to the volumetric flow rate of the entering stream. The concentration of substrate in the aqueous phase within the reactor is equal to s_{98} for a well-stirred system. Thus, equation (L) becomes

V \left(s_{\text {feed }}-s_{98}\right)=\mu X =\left(\frac{\mu_{\max } s_{98}}{K_S+s_{98}}\right) x V_R         (M)

From a physical interpretation of this relation, it becomes obvious that if the amount of biomass in the reactor is to grow the concentration of substrate in the feed stream must exceed the concentration of substrate in the growth medium. Otherwise, the constraints on the fed batch operation would not be satisfied with respect to the desire for a constant substrate concentration during the fed batch period. As the concentration of substrate in the feed stream increases, the smaller the volumetric flow rate of the feed stream can be and the longer will be the time necessary to exceed the volumetric capacity of the bioreactor. There are multiple possibilities for the combinations of dilution rate and the feed concentration of substrate that will satisfy equation (M). The upper limit on the substrate concentration is established by its solubility in the growth medium. In terms of equation (M) the volumetric flow rate of the entering stream would be selected so that the product V(S_{feed} − S_{98}) is equal to the rate of consumption of substrate given by the Monod equation. Thus, at the time use of the feed stream is initiated, equation (M) indicates
that

V [(50)-(0.70)]=\left[\frac{3.16(0.70)}{2.78+0.70}\right](26.82)         (N)

or V = 0.346 L/day = 0.0144 L/h. This flow rate will need to be increased as the amount of time the additional substrate is fed increases. An increased feed rate is necessary to offset the increase in the rates of consumption of substrate and production of additional biomass, as well as dilution effects caused by the increase in the fluid volume within the bioreactor. The numbers above indicate that the daily change in volume is a very significant fraction of the working volume of the bioreactor (1 L). Nonetheless, the volume change will not prevent us from solving this problem provided that there is sufficient head space within the bioreactor to accommodate the increase in volume of the aqueous phase.

Equation (I) describes the rate at which biomass is being produced during the fed- batch period. It can also be written as

\frac{d X }{d t}=Y_{ X / S }\left(\frac{\mu_{\max } s_{98}}{K_S+s_{98}}\right) x V_R           (O)

Separation of variables and integration subject to the initial condition that at the time the fed-batch phase of the operation begins, X = X_{98} gives

\int_{X_{98}}^X \frac{d X }{ X }=\int_{21.47}^t Y_{ X / S }\left(\frac{\mu_{\max } s_{98}}{K_S+s_{98}}\right) d t             (P)

or

\ln \left(\frac{ X }{ X _{98}}\right)=Y_{ X / S }\left[\frac{\mu_{\max } s_{98}}{K_S+s_{98}}\right](t-21.47)                   (Q)

Substitution of numerical values gives

\ln \left(\frac{ X }{26.82}\right)=0.709\left[\frac{(3.16 / 24)(0.70)}{2.78+(0.70)}\right](t-21.47) = 0.01878(t − 21.47)                (R)

If fed-batch operation is employed for a total of 48 h (t = 69.47 h), equation (R) indicates that

X = 66.06 g                   (S)

The amount of biomass produced via the combination of batch and fed-batch operation is 2.46 times as great as that produced in the strictly batch portion of the fermentation. However, the total time involved in the fed-batch operation is greater than that for strict batch operation by a factor of 69.47/28.07 = 2.47. Hence, there does not appear to be a major advantage in terms of biomass production per unit time in going to a fed-batch mode of operation for the second stage of production. Nonetheless, examination of Figure I13.4 indicates that application of the fed-batch mode of operation produces substantial amounts of biomass over and above that produced during the strictly batch portion of the reaction. Analysis of the slopes of the biomass plot in the figure during the two stages of the reaction indicates that the reaction rate is fastest during the last portion of the strictly batch stage of the process cycle.

Readers should note that for purposes of process control the flow rate of the feed stream can be determined at any time using equation (M) and the value of the total weight of biomass in the bioreactor at that time. When the fed batch phase of operation is initiated, the requisite feed rate is 14.4 mL/h, and at the end of this phase the necessary feed rate is 35.5 mL/h.