Determination of the Holding Time Necessary to Achieve 98% Conversion of the Limiting Substrate in a Batch Fermentation Consider the fermentation of a representative microorganism in a well-agitated laboratory-scale bioreactor with an initial working volume of 1 L and sufficient headspace to

Question

Determination of the Holding Time Necessary to Achieve 98% Conversion of the Limiting Substrate in a Batch Fermentation

Consider the fermentation of a representative microorganism in a well-agitated laboratory-scale bioreactor with an initial working volume of 1 L and sufficient headspace to accommodate any volume changes that accompany the biochemical reactions taking place. (See also Illustration 13.4 concerning fed-batch operation of this reactor.) Solute S is the substrate species that limits the growth of the microorganism. Immediately after inoculation of the growth medium, the concentrations of solute S and the microorganism are 35 and 2.5 g/L, respectively. Data from previous experimental trials indicate that the yield coefficient for the microorganism is substantially constant over the course of the experiment: [latex]Y_{X/S} = 0.709[/latex].

The rate at which the microorganism grows obeys a Monod rate expression with a maximum specific growth rate of 3.16 day

^{−1}

. We are asked to ascertain the manner in which the rate at which soluble substrates are transformed to biomass depends on the half-saturation constant by considering the effect of variations in the value of this parameter on the time necessary to achieve a specified level of consumption of the limiting substrate. In particular, consider
the times necessary to achieve 98% consumption of the limiting substrate for

K_{S}

values of 0.278, 2.78, and 27.8 g/L. These values of the half-saturation constant cover the range of conditions from situations in which

K_{S}

is small compared to the concentration of the limiting substrate to those in which

K_{S}

is comparable in magnitude to the initial concentration of substrate. The largest parameter value is quite large compared to those typically encountered in industrialscale biotransformations.

Determine the holding time in the bioreactor necessary to accomplish the conversion specified and the total cycle time for each batch. To facilitate the analysis for this illustrative example, you may presume that the working volume of the growth medium remains constant because the volume of the incoming solution used to control the pH is controlled to exactly offset the volume change what would otherwise accompany the transformation of soluble substrates into biomass. The sum of the nonproductive times during the cycle for a single experimental trial is 6.6 h, of which 3.5 h is associated with the lag period for growth of the microorganism.

Accepted Answer

The design equation for a batch bioreactor is a variation of equation (8.1.5).

[latex]t_2-t_1=N_{ A 0} \int_{f_{ A 1}}^{f_{ A 2}} \frac{d f_{ A }}{\left(-r_{ A } ight) V_R}[/latex] (8.1.5)

This equation can be expressed in terms of the total mass of the microorganism, X, in the bioreactor as

[latex]\frac{d X }{d t}=\mu X[/latex] (A)

where the specific growth rate is of the Monod form,

[latex]\mu=\frac{\mu_{\max } s}{K_S+s}[/latex] (B)

and where s is the concentration of the substrate in the aqueous growth medium. Combination of equations (A) and (B) followed by integration yields

[latex]t_{ ext {batch }}=\int_{X_0}^X \frac{\left(K_S+s ight) d X }{\mu_{\max } s X } =\frac{1}{\mu_{\max }}\left(\int_{X_0}^X \frac{K_S d X }{s X }+\int_{X_0}^X \frac{d X }{ X } ight)[/latex] (C)

The variables s and X can be related using the overall yield coefficient

[latex]V_R\left(s_0-s ight)=\frac{ X - X _0}{Y_{ X / S }}[/latex] (D)

or

[latex]s=s_0-\frac{ X - X _0}{V_R Y_{ X / S }}[/latex] (E)

Combination of equations (C) and (E) gives

[latex]\mu_{\max } t_{ ext {batch }}=K_S \int_{X_0}^X \frac{d X }{ X \left\{s_0-\left[\left( X - X _0 ight) /\left(V_R Y_{ X / S } ight) ight] ight\}}+\int_{X_0}^X \frac{d X }{ X }[/latex] (F)

Integration gives

[latex]\mu_{ ext {max }} t_{ ext {batch }}=\frac{K_S}{s_0+\left[ X _0 /\left(V_R Y_{ X / S } ight) ight]} imes \ln \left[\frac{ X }{s_0-\left[\left( X - X _0 ight) /\left(V_R Y_{ X / S } ight) ight]} ight]_{X_0}^X+\ln \left[\frac{ X }{ X _0} ight]=\left[\frac{K_S}{s_0+\left[ X _0 /\left(V_R Y_{ X / S } ight) ight]}+1 ight] \ln \left[\frac{ X }{ X _0} ight]-\left[\frac{K_S}{s_0+\left[ X _0 /\left(V_R Y_{ X / S } ight) ight]} ight] imes\ln \left[\frac{\left.s_0-\left[ X - X _0 ight] /\left(V_R Y_{ X / S } ight) ight]}{s_0} ight][/latex] (G)

Equation (G) may be simplified using equation (E) to obtain

[latex]\mu_{\max } t_{ ext {batch }}=\left[\frac{K_S}{s_0+\left[ X _0 /\left(V_R Y_{ X / S } ight) ight]}+1 ight] imes\ln \left(\frac{ X }{ X _0} ight)-\left[\frac{K_S}{s_0+\left[ X _0 /\left(V_R Y_{ X / S } ight) ight]} ight] \ln \left(\frac{s}{s_0} ight)[/latex] (H)

or

[latex]t_{ ext {batch }}=\frac{1}{\mu_{\max }}\left\{\left[\frac{K_S}{s_0+\left[ X _0 /\left(V_R Y_{ X / S } ight) ight]} ight] imes ight. \left.\ln \left(\frac{ X s_0}{ X _0 s} ight)+\ln \left(\frac{ X }{ X _0} ight) ight\}[/latex] (I)

Moreover, the cycle time per batch is the sum of the productive and nonproductive periods:

[latex]t_{ ext {cycle }}=t_{ ext {batch }}+t_{ ext {nonproductive }}[/latex] (J)

At 98% conversion of the limiting substrate,

[latex]( S )=\left(1-f_S ight)\left( S _0 ight)=(1-0.98)(35)=0.70 g[/latex] (K)

From equation (D) one can determine the amount of biomass formed at the time 98% of the limiting substrate has been consumed:

[latex]X = X _0+Y_{ X / S 0}\left[s_0-s ight]=2.5+0.709[35-0.70] = 26.82 g[/latex] (L)

If we had waited until the time at which no more biomass was being formed (when s = 0) we would find that regardless of the value of [latex]K_{S} , X_{∞} = 27.32 g[/latex].

At this point we have all the pieces necessary to ascertain the fermentation times and cycle times corresponding to the three values of the half-saturation constant indicated in the problem statement. These calculations are readily accomplished using a spreadsheet based on equations (H)–(K). The results are summarized in the following table.

[latex]K_{S}[/latex] (g/L)	[latex]\frac{K_S}{s_0+\left[ X _0 /\left(V_R Y_{ X / S } ight) ight]}[/latex]	[latex]t_{batch}[/latex] (h)	[latex]t_{cycle}[/latex] (h)
0.278	0.00722	18.37	24.97
2.78	0.0722	21.47	28.07
27.8	0.722	52.48	59.08

Inspection of the spreadsheet from which the values were derived indicates that the total quantity of microorganism produced at 98% conversion of the substrate in this strictly batch process is equal to 26.82 g. Equations (I) to (L) can be utilized in a spreadsheet format to prepare plots of the masses of cells and substrate in the bioreactor versus the holding time. Examination of the entries in the table indicates that as [latex]K_{S}[/latex] decreases, the time necessary to achieve 98% conversion of the limiting substrate decreases. The most significant decrease in absolute terms is observed when transitioning from the condition in which [latex]K_{S}[/latex] is comparable to the initial value of the substrate concentration to the condition in which [latex]K_{S}[/latex] is comparable to the value the substrate concentration at ca. 92% consumption. Figure I13.3 depicts the time courses of the masses of substrate and cells in the bioreactor for the latter case in which [latex]K_{S}[/latex] is 2.78 g/L.

Question 13.3: Determination of the Holding Time Necessary to Achieve 98% C...

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