# Continuous stirred-tank reactor (CSTR) in parallel problem

I am having a difficult time solving the following problem. The only thing we learned in class was parallel continuous stirred-tank reactor (CSTR) with equal-sized reactors and equal flow rates for each reactor (which I think also implies the volumetric flow rate is the same across all reactors).

When the 1st order liquid-phase reaction $$A \rightarrow B$$ (elementary) is carried out in a single CSTR (initial concentration of species A = $$C_{A0}$$, reactor volume = $$V_R$$, volumetric flow rate = $$\nu_0$$), one can achieve 80% conversion. The same reaction is to be conducted in two CSTRs connected in parallel while keeping the total reactor volume constant at $$V_R$$. How should we split the volumetric flow rate when the final conversion ($$X_{final}$$) is 0.786.

Initial Case Applying a mole balance to the single CSTR in steady state yields \begin{align} v_0C_{A0} - v_0C_{A} - kC_{A}V &= 0 \\ C_{A} &= \frac{v_0C_{A0}}{v_0 + kV} \tag{1} \end{align} The conversion, combining with Eq. (1), is \begin{align} X_1 &= \frac{C_{A0} - C_A}{C_{A0}} \\ X_1 &= \frac{C_{A0} - \dfrac{v_0C_{A0}}{v_0 + kV}}{C_{A0}} \\ X_1 &= 1 - \frac{v_0}{v_0 + kV} \\ X_1 &= \frac{kV}{v_0 + kV} \rightarrow kV = \frac{X_1v_0}{1 - X_1} \tag{2} \end{align} We will use Eq. (2) later.
Second Case Now the volumetric flow rate splits in two. Let me call that value in the upper branch $$\alpha v_0$$, and the one for the lower branch $$(1 - \alpha)v_0$$. Since we have a bifurcation, the concentration of reactant $$A$$ doesn't change, so the value is still $$C_{A0}$$ when entering both reactors.
We denote the exit concentration in the reactor at the top $$C_{A1}$$, and at the bottom $$C_{A2}$$. Applying a mole balance to both CSTR's in steady state, with equal volumes but of value $$V/2$$, yields \begin{align} \alpha v_0C_{A0} - \alpha v_0 C_{A1} &- (k C_{A1})\frac{V}{2} = 0 \\ (1 - \alpha)v_0C_{A0} - (1 - \alpha)v_0C_{A2} &- (kC_{A2})\frac{V}{2} = 0 \\ C_{A1} = \frac{\alpha v_0C_{A0}}{\alpha v_0 + kV/2} &\hspace{0.25 cm} C_{A2} = \frac{(1 - \alpha)v_0C_{A0}}{(1 - \alpha)v_0 + kV/2} \tag{3,4} \\ \end{align} Now, upon mixing the streams, the volumetric flow rate recovers its value $$v_0$$. However, the concentration changes, and we denote its value by $$C_{A3}$$. This one is obtained by a mole balance at steady state, combining with Eqs. (3) and (4) \begin{align} \alpha v_0C_{A1} + (1 - \alpha)v_0C_{A2} &= v_0 C_{A3} \\ C_{A3} &= \alpha C_{A1} + (1 - \alpha)C_{A2} \\ C_{A3} &= \frac{\alpha^2 v_0C_{A0}}{\alpha v_0 + kV/2} + \frac{(1 - \alpha)^2v_0C_{A0}}{(1 - \alpha)v_0 + kV/2} \tag{5} \\ \end{align} The conversion, using Eq. (5), yields \begin{align} X_2 &= \dfrac{C_{A0} - C_{A3}}{C_{A0}} \\ X_2 &= \dfrac{C_{A0} - \dfrac{\alpha^2 v_0C_{A0}}{\alpha v_0 + kV/2} - \dfrac{(1 - \alpha)^2v_0C_{A0}}{(1 - \alpha)v_0 + kV/2}}{C_{A0}} \\ X_2 &= 1 - \frac{\alpha^2 v_0}{\alpha v_0 + kV/2} - \frac{(1 - \alpha)^2v_0C_{A0}}{(1 - \alpha)v_0 + kV/2} \tag{6} \\ \end{align} Now we combine Eq. (2) with Eq. (6) \begin{align} X_2 &= 1 - \frac{\alpha^2 v_0}{\alpha v_0 + \dfrac{X_1v_0}{2(1 - X_1)}} - \frac{(1 - \alpha)^2v_0}{(1 - \alpha)v_0 + \dfrac{X_1v_0}{2(1 - X_1)}} \\ X_2 &= 1 - \frac{\alpha^2}{\alpha + \dfrac{X_1}{2(1 - X_1)}} - \frac{(1 - \alpha)^2}{(1 - \alpha) + \dfrac{X_1}{2(1 - X_1)}} \tag{7} \\ \end{align} Eq. (7) is a non-linear equation that we need to solve. Note that if $$\alpha = 1/2$$, the left-hand side yields 0.8, which is reasonable.
It is informed to us that $$X_2 = 0.786$$ and that the original conversion was $$X_1 = 0.8$$. Upon solving numerically, we found actually two possible solutions $$\boxed{\alpha \in (0.33500,0.66500)}$$ However, since both branches are "identical" to each other, it is the same if 33.5% goes up, and 66.5% goes down, or vice versa.