# Operating a chemostat at the optimum dilution rate [closed]

I am wondering what precautions should one take to operate a chemostat at the optimal dilution rate. This is a question on one of the exercise sheets from a class I took two semesters ago and that I am currently revising, but I cannot really make much progress on this. I tried looking it up in the literature, but with no result, so I would really appreciate some help with this.

EDIT: Here dilution refers to the dilution with a sterile broth medium and it equals the ratio of the flow divided by the volume of the chemostat.

• What sort of "dilution rate" are you speaking of ? Jun 28, 2023 at 19:41
• @Maurice I added some explanations, is it clear now? Jun 28, 2023 at 19:55

Paradoxically, the optimum operation is not recommended for a chemostat, so typically it should be avoided. This is because the optimum dilution rate, more often than we would like, is near the critical dilution rate. Therefore, small variations of the volumetric flow rate $$F$$ will lead to a washing of the biomass. I will present to you the proof of this behaviour, and then use some data that I personally used in my biochemistry classes, that illustrates this problem.

The sketch of an academic chemostat is on the left, while reality is on the right:

1. Biomass balance With a zero inlet flow of biomass, i.e. $$X_0 = 0$$, at steady state we have \begin{align} \require{cancel} \cancel{FX_0} - FX + \mu XV &= \frac{\mathrm{d}X}{\mathrm{d}t} = 0 \\ -FX + \mu XV &= 0 \\ -F + \mu V &= 0 \rightarrow \boxed{\mu = \frac{F}{V} = D} \tag{1} \\ \end{align}

2. Substrate balance Let $$Y_\mathrm{X/S}$$ denote the stoichiometric/maximum yield of biomass/substrate. At steady state, we have \begin{align} \require{cancel} FS_0 - FS - \frac{\mu X}{Y_\mathrm{X/S}}V &= \frac{\mathrm{d}S}{\mathrm{d}t} = 0 \\ F(S_0 - S) - \frac{\mu X}{Y_\mathrm{X/S}}V &= 0 \hspace{1 cm} (\text{divide by V)} \\ D(S_0 - S) - \frac{\mu X}{Y_\mathrm{X/S}} &= 0 \hspace{1 cm} (\text{use Eq. (1))} \\ \cancel{D}(S_0 - S) - \frac{\cancel{D} X}{Y_\mathrm{X/S}} &= 0 \\ (S_0 - S) - \frac{X}{Y_\mathrm{X/S}} &= 0 \rightarrow X = Y_\mathrm{X/S}(S_0 - S) \tag{2} \end{align} Assuming the specific growth rate follows the Monod's equation \begin{align} \mu &= \frac{\mu_\mathrm{max}S}{K_\mathrm{S} + S} \\ S &= \frac{\mu K_\mathrm{S}}{\mu_\mathrm{max} - \mu} \hspace{1 cm} (\text{use Eq. (1))} \\ S &= \frac{D K_\mathrm{S}}{\mu_\mathrm{max} - D} \tag{3} \\ \end{align} and combining Eq. (2) and (3) $$X = Y_\mathrm{X/S}\left(S_0 - \frac{D K_\mathrm{S}}{\mu_\mathrm{max} - D}\right) \tag{4}$$ Eq. (4) shows a feature of the chemostat at steady state. If you don't choose well the volumetric flow rate, the biomass at the exit will be zero (the washing I mentioned at the beginning). Setting Eq. (4) to zero gives the critical dilution rate $$\boxed{D_\mathrm{crit} = \frac{\mu_\mathrm{max} S_0}{K_\mathrm{S} + S_0}} \tag{5}$$

3. Optimum condition The biomass productivity is defined as $$P_\mathrm{X} := DX$$. Thus, combining Eqs. (1) and (4) and differentiating with respect to the dilution rate $$D$$ (I let that derivative to you) gives the optimum dilution rate \begin{align} P_\mathrm{X} &= D Y_\mathrm{X/S}\left(S_0 - \frac{D K_\mathrm{S}}{\mu_\mathrm{max} - D}\right) \\ \frac{\mathrm{d}P_\mathrm{X}}{\mathrm{d}D} &= \frac{\mathrm{d}}{\mathrm{d}D}\left[D Y_\mathrm{X/S}\left(S_0 - \frac{D K_\mathrm{S}}{\mu_\mathrm{max} - D}\right)\right] = 0 \rightarrow \boxed{D_\mathrm{opt} = \mu_\mathrm{max} \left(1 - \sqrt{\frac{K_\mathrm{S}}{K_\mathrm{S} + S_0}}\right)} \tag{6} \end{align}

4. Example In my class we used saccharomyces cerevisiae at $$T = 30 ^\circ \pu{C}$$ and the limiting nutrient was glucose. The important data is: $$\begin{array}{|c|c|} \text{Magnitude} & \text{Value} \\ \hline \mu_\mathrm{max} & \pu{0.5 h^-1} \\ K_\mathrm{S} & \pu{0.025 g glucose/L} \\ S_0 & \pu{12 g glucose/L} \\ Y_\mathrm{X/S} & \pu{0.5 g biomass/g glucose} \\ \end{array}$$ Replacing this values in Eqs. (5) and (6) give $$$$D_\mathrm{opt} = \pu{0.4772 h^-1} \hspace{1 cm} D_\mathrm{crit} = \pu{0.4990 h^-1}$$$$ This goes back when I started the post, both values are pretty near to each other. The evolution of $$X$$, $$S$$, and $$P_\mathrm{X}$$ is shown below:

If the dilution rate goes up a bit, you abruptly wash all the biomass. If it lowers a bit, the consequences are not so serious. The golden characteristic of the chemostat is depicted in Eq. (1), because we can manipulate specific growth rate by changing $$D$$. Unfortunately, by the same cause, changes in $$D$$ will bring up a new set of steady-state values (after a transient period), that may be vastly different.

Fortunately, it can be proven that the chemostat is an autoregulator, and a new steady state will always be settled. I invite you to search for the Monod parameters for various microorganisms, apply these equations, and see that it is generally the case.