Difference between revisions of "Team:Heidelberg/Model/Lagoon Contamination"

Contaminations with bacteria other than the host cells that are part of the experiment, funghi or other microorganisms can survive in the lagoons, it is necessary to take precautions. Since these can be costly (e.g. fungicides) an simulation of contamination growth can save resources. If the constant dilution rate of the lagoons is sufficiently high that the overall change in population size of the contamination is negative, the contamination will wash out and not be a lasting problem. Note: contamination with viruses such as bacteriophages is not modelled here, because exponential growth of the contamination is assumed. This is plausible for bacteria and funghi since the lagoons are constantly diluted so that there is always fresh medium and space available. The change in concentration of the contamination can be described as $$ \frac{\partial c_{X}(t)}{\partial t} = -\Phi_{L} \cdot c_{X}(t) + \frac{\ln{2} }{T_{X} \cdot c_{X}(t) $$ There is the dilution term containing the factor \(\Phi_{L}\) that is the flow rate trough the lagoon in volumes per hour and the growth term of the contamination based on the contaminations generation time \(t_{X}\). Solving this equation yields $$ c_{X}(t) = c_{X}(t_{0} \cdot e^{\big(-\Phi_{L} + \frac{\ln{2} }{T_{X} }\big) \cdot t} $$ Here \(c_{X}(t_{0}\) is introduced, which is the initial concentration of the contamination. The following statements can be made:
The contamination expands if the flow rate \(\Phi_{L}\) is greater than the growth factor \(\frac{ln(2)}{t_{X} }\), $$ c_{X}(t) > 0, \quad if \Phi_{L} > \frac{ln(2)}{t_{X} } $$ it remains constant if the flow rate is exactly the growth factor $$ c_{X}(t) = 0, \quad if \Phi_{L} = \frac{ln(2)}{t_{X} } \end{cases} $$ and it diminishes, when the flow rate is higher than the growth factor. $$ c_{X}(t) \less 0, \quad if \Phi_{L} \less \frac{ln(2)}{t_{X } } \end{cases} $$