We have this conformation of connected water tanks: We can model the first tank as: \[ \begin{cases} \frac{dm}{dt} = \dot{m}_{in} - \dot{m}_{out} \\ \dot{m}_{out}=\sqrt{m} \\ \dot{m}_{in}=10 \\ m(t=0)=0 \end{cases} \] Where \(m\) is the mass inside the tank, \( \dot{m}_{in}\) is the mass entering the tank and \( \dot{m}_{out}\) is the mass leaving the tank. Units: Kg for mass, minutes for time. Tip: \(\int_{0}^{y} \frac{1}{a - \sqrt{x}} \, dx = 2a \ln \left| \frac{a}{a - \sqrt{y}} \right| - 2\sqrt{y}\) Take a look at the following statements: 1. The time for the first tank to achieve 80% of it's steady-state mass would be roughly 27 minutes. 2. Assuming the second tank follow the same model from the first tank, but it's inlet is defined by the outlet of the first tank, after a very long time, the mass inside the tank would be roughly 10 Kg. 3. Assume that at a given time, the entire tank system behaves like the following flow network: The residual network related to the syste...
