Q7RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

For the interconnected tanks problem of Section  5.1, page  241 , suppose that instead of pure water being fed into the tank A, a brine solution with concentration $\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{ }\mathbf{kg}\mathbf{/}\mathbf{L}$ is used; all other data remain the same. Determine the mass of salt in each tank at time  $\mathit{t}$ if the initial masses are   and ${\mathbf{y}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{ }\mathbf{kg}$.

Step-by-Step Solution

Answer

The change of mass of salt n tanks A  and  B are:

 $$\begin{gathered} \mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{225}{\mathbf{e}}^{\mathbf{-}\mathbf{t}\mathbf{/}\mathbf{2}}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{475}{\mathbf{e}}^{\mathbf{-}\mathbf{t}\mathbf{/}\mathbf{6}}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{8} \\ \mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{45}{\mathbf{e}}^{\mathbf{-}\mathbf{t}\mathbf{/}\mathbf{2}}\mathbf{-}\mathbf{6}\mathbf{.}\mathbf{95}{\mathbf{e}}^{\mathbf{-}\mathbf{t}\mathbf{/}\mathbf{6}}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{8} \end{gathered}$$

 

Step 1: Finding the change of concentration of the tank A,B

Let’s first derive a system of differential equations that describes the change of salt in each tank at a time t. One knows that $\mathbf{dx}\mathbf{/}\mathbf{dt}\mathbf{=}\mathbf{inputrate}\mathbf{ }\mathbf{-}\mathbf{outputrate}$  and in the tank t,   two pipes bring salt in it, the left one at the rate   $\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{ }\mathbf{kg}\mathbf{/}\mathbf{L}\mathbf{\times }\mathbf{6}\mathbf{ }\mathbf{L}\mathbf{/}\mathbf{min}$ , and the right lower pipe brings salt at the rate $\mathbf{2}\mathbf{ }\mathbf{L}\mathbf{/}\mathbf{min}\mathbf{\times }\mathbf{y}\mathbf{/}\left(\mathbf{24}\mathbf{L}\right)$. The upper pipe carries salt out of the tank A  at the rate of $2\mathbf{L}\mathbf{/}\mathbf{min}\mathbf{\times }\mathbf{x}\mathbf{/}\mathbf{(}\mathbf{24}\mathbf{ }\mathbf{L}\mathbf{)}$. So, one has that the change of the concentration of salt in the tank  A is $\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{\times }\mathbf{6}\mathbf{+}\mathbf{2}\frac{\mathbf{y}}{\mathbf{24}}\mathbf{-}\mathbf{8}\frac{\mathbf{x}}{\mathbf{24}}$.

 

One has that the upper pipe brings salt into the tank  B by the rate of  $\mathbf{8}\mathbf{ }\mathbf{L}\mathbf{/}\mathbf{min}\mathbf{\times }\mathbf{x}\mathbf{/}\mathbf{(}\mathbf{24}\mathbf{ }\mathbf{L}\mathbf{)}$, the lower pipe carries salt out of tank B by the rate  $\mathbf{2}\mathbf{ }\mathbf{L}\mathbf{/}\mathbf{min}\mathbf{\times }\mathbf{y}\mathbf{/}\left(\mathbf{24}\mathbf{L}\right)$ and the right pipe carries salt out by the rate of $\mathbf{6}\mathbf{ }\mathbf{L}\mathbf{/}\mathbf{min}\mathbf{\times }\mathbf{y}\mathbf{/}\mathbf{(}\mathbf{24}\mathbf{ }\mathbf{L}\mathbf{)}$ so the change of concentration of salt in the tank  B is $\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{8}\frac{\mathbf{x}}{\mathbf{24}}\mathbf{-}\mathbf{2}\frac{\mathbf{y}}{\mathbf{24}}\mathbf{-}\mathbf{6}\frac{\mathbf{y}}{\mathbf{24}}$

Step 2: Substituting the values

So, to determine the mass of salt in each rank one has to solve the following system:

 $$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\frac{\mathbf{x}}{\mathbf{3}}\mathbf{+}\frac{\mathbf{y}}{\mathbf{12}}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{2} \\ \frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\frac{\mathbf{y}}{\mathbf{3}}\mathbf{+}\frac{\mathbf{x}}{\mathbf{3}} \end{gathered}$$

The second equation gives us that

 $$\begin{gathered} \frac{\mathbf{x}}{\mathbf{3}}\mathbf{=}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{+}\frac{\mathbf{y}}{\mathbf{3}} \\ \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{3}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{dy}}{\mathbf{dt}} \end{gathered}$$

Substituting this into the first equation of the system one will get

 $$\begin{gathered} \mathbf{3}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{-}\frac{\mathbf{y}}{\mathbf{3}}\mathbf{+}\frac{\mathbf{y}}{\mathbf{12}}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{2} \\ \Rightarrow \mathbf{3}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{t}}^{\mathbf{2}}}\mathbf{+}\mathbf{2}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{+}\frac{\mathbf{y}}{\mathbf{4}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{2} \\ \Rightarrow \mathbf{12}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{+}\mathbf{8}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{8} \end{gathered}$$