Consider the steady state material balance equations for the first chemostat. These relations are the same as those given by equations (13.2.71) and (13.2.76)
s_1=\frac{K_{S 1} D_1}{\mu_{\max 1}-D_1} (13.2.76)
for the biomass and limiting substrate, respectively. However, for the second chemostat we must reformulate the steady state material balances even though they are of the usual form:
input = output + disappearance by reaction
x_1 \nu+0=\left(\nu+\nu^*\right) x_2-\mu_2 x_2 V_{R 2} (A)
where we have assumed that the volumetric flow rate of the effluent from the second reactor is equal to the sum of the flow rates of the two incoming streams. Algebraic manipulation of this equation yields
\mu_2=\frac{\nu+\nu^*}{V_{R 2}}-\frac{x_1 \nu}{x_2 V_{R 2}} (B)
where the first term on the right of equation (B) can be interpreted as a dilution rate based on the volumetric flow rate of the effluent from the second chemostat. The biomass specific growth rate in the second chemostat can be expressed in terms of the Monod equation as
\mu_2=\frac{\mu_{\max 2} s_2}{K_{S 2}+s_2} (C)
where the kinetic parameters are evaluated at the environmental conditions prevailing in the second chemostat. The corresponding material balance on the limiting substrate for steady-state operation of the second chemostat is
s_1 \nu+s_0^* \nu ^*=\left( \nu + \nu ^*\right) s_2+\frac{\mu_2 x_2 V_{R 2}}{Y_{ X / S }} (D)
In this situation the equation for the concentration of biomass in the effluent from the second chemostat becomes
x_2=\frac{Y_{ X / S }\left[\left(s_1-s_2\right) \nu +\left(s_0^*-s_2\right) \nu ^*\right]}{\mu_2 V_{R 2}} (E)
At this point one can employ machine computation to determine the compositions of the effluent streams from each of the two chemostats using equations (13.2.71),(13.2.76), (B), (C), and (E), together with specified values of appropriate process parameters.
