The maximum reaction rate corresponds to a condition for which the derivative of the rate law with respect to the concentration of biomass is zero.
0=\frac{d r_x}{d x}=\mu_{\max } \frac{d\left\{\left[1-\left(x / x_{\infty}\right)\right] x\right\}}{d x}=\left(1-\frac{x}{x_{\infty}}\right)+x\left(\frac{-1}{x_{\infty}}\right) (B)
Thus, at the maximum rate,
x=\frac{x_{\infty}}{2}=\frac{140}{2}=70 g / L (C)
The corresponding rate is
r_{x 1}=1.2\left(1-\frac{70}{140}\right) 70=42 g /( L \cdot \text { day }) (D)
A steady state material balance on the biomass around the first reactor in the cascade yields
input = output + disappearance by reaction (E)
0=x_1 \nu-r_{x 1} V_{R 1} (F)
where we have designated the concentration of biomass leaving the first CSTBR as x_{1}. The corresponding rate of production of biomass per unit volume of this reactor is r_{x1}. Rearrangement of equation (F) and introduction of the dilution rate (D_{1} = \nu/V_{R1}) gives
r_{x 1}=x_1 \frac{\nu}{V_{R 1}}=D_1 x_1 (G)
Equation (G) indicates that a plot of r_{1} versus x_{1} representing the material balance on the first CSTBR will be a straight line linking the origin and the point x_{1} , r_{x1}. The slope of this line is the dilution rate for the first CSTBR (D_{1}). The intersection of this straight line and the graphical representation of the logistic rate law (expressed per unit volume of growth medium) will correspond to steady-state conditions in the first CSTBR. To minimize the volume of the first reactor, we specified operation of this reactor at the maximum in the plot of the rate versus biomass concentration curve (see Figure I13.6): namely, at x = 70 g/L and r_{x1} = 42 g/(L⋅day). The value of the dilution rate is equal to the slope of the line linking the origin and the point of intersection of the straight line with the logistic rate curve: namely, 0.6 day−1. Thus, from equation (G),
V_{R 1}=\frac{\nu}{D_1}=\frac{4000}{0.6}=6667 L (H)
The volume of the first CSTBR is 6667 L.
The volumes of the second and third reactors are each 50% larger than the volume of the previous reactor. Hence,
V_{R 2}=1.5 V_{R 1}=1.5(6667)=10,000 L (I)
and
V_{R 3}=1.5 V_{R 2}=1.5(10,000)=15,000 L (J)
The dilution rates corresponding to these reactor volumes are
D_2=\frac{\nu}{V_{R 2}}=\frac{4000}{10,000}=0.4 day ^{-1} (K)
D_3=\frac{\nu}{V_{R 3}}=\frac{4000}{15,000}=0.26667 day ^{-1} (L)
Material balances around the second and third CSTBRs lead to the following analogs of equation (F):
r_{x 2}=D_2\left(x_2-x_1\right) for the second CSTBR (M)
and
r_{x 3}=D_3\left(x_3-x_2\right) for the third CSTBR (N)
The straight lines describing the material balances on the second and third CSTBRs [equations (M) and (N), respectively] can be drawn through the point x = 70, r = 0 with slope = 0.4 for the second reactor, and through the point x = 120.45, r = 0 with slope 0.26667 for the third reactor (see Figure I13.6). The intersections of these straight lines with the curve representing the rate law indicate the following values for biomass concentrations and rates of reaction.
\begin{array}{ll}x_2=120.45 g / L & x_3=136.37 g / L \\ \\ r_2=20.18 g /( L \cdot \text { day }) & r_3=4.245 g /( L \cdot \text { day })\end{array}
We have listed above more significant figures than one would expect for a graphical solution. The excess was determined using the spreadsheet on which Figure I13.6 is based. One could obtain similar results using engineering software containing an equation solver and equations (I) to (N).
Readers may wish to compare and contrast the methods of analysis and results of this illustrative example with the corresponding aspects of Illustration 8.7. This example employed a graphical approach to analyze the behavior of a cascade of three reactors employed to conduct a reaction whose rate increased continuously with reactant concentration.
