Consider Example 16.1, where we demonstrated that two bacteria competing for a single nutrient in a chemostat (well-mixed) could not coexist. Consider the situation where $B$ can adhere to a surface but $A$ cannot. Redo the balance equations, where $a$ is the surface area available per unit reactor volume and the rate of attachment is first order in $X_{B}$ with a rate constant $k_{a B}$. The sites available for attachment will be $\left(X_{B M}^{A t}-X_{B}^{A t}\right) a V$. The attached cells can detach with a first-order dependence on the attached cell concentration $\left(X_{B}^{A t}\right)$ with a rate constant of $k_{d B}$. Attached cells grow with the same kinetics as suspended cells.
a. Without mathematical proofs, do you think coexistence may be possible? Why or why not?
b. Consider the specific case below and solve the appropriate balance equations for $D=0.4 \mathrm{~h}^{-1}$ :
$$
\begin{aligned}
\mu_{m A} &=1.0 \mathrm{~h}^{-1} ; \quad \mu_{m B}=0.5 \mathrm{~h}^{-1} \\
K_{S A} &=K_{S B}=0.01 \mathrm{~g} / \mathrm{l} \\
Y_{A / S} &=Y_{B / S}=0.5 \mathrm{~g} / \mathrm{g} \\
S_{0} &=5 \mathrm{~g} / 1, \quad a=10 \mathrm{~cm}^{2} / \mathrm{cm}^{3} \\
X_{B M}^{A t} &=1 \times 10^{-4} \mathrm{~g} / \mathrm{cm}^{2} \\
k_{d B} &=0.5 \mathrm{~h}^{-1}, \quad k_{a B}=1000 \mathrm{~cm}^{3} / \mathrm{g}-\mathrm{h}
\end{aligned}
$$