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Consider a large tank holding 1000 L of pure water into which a brine solution of salt beings to flow at a rate of 6 L/min. The solution inside the tank is kept well stirred and is flowing out of the tank at a rate of 6 L/min. If the concentration of salt in the brine entering the tank is 0.1 kg/L, determine when the concentration of salt in the tank will reach 0.05 kg/L.​

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[tex] \Large \mathbb{SOLUTION:} [/tex]

[tex] \begin{array}{l} \bold{Given:}\ \begin{cases} \ V(0) = 1000 \\ \textsf{Input of brine} = 6\ \textsf{L/min} \\ C_{\textsf{in}}= 0.01\ \textsf{kg/L} \\ \textsf{Output of brine} = 6\ \textsf{L/min} \end{cases} \\ \\ \textsf{Let }y(t)\textsf{ be the amount of salt dissolved in the} \\ \textsf{tank (in kg) after }t\textsf{ minutes. Now, we have }\\ C_{\textsf{out}} = \dfrac{y(t)}{1000 + 6t - 6t} = \dfrac{y(t)}{1000} = x(t), \\ \textsf{where }x(t)\textsf{ is the salt concentration in the brine} \\ \textsf{leaving the tank at some time }t. \\ \\ \textsf{The basic principle determining the differential} \\ \textsf{equation is} \\ \\ \Large \quad \quad \dfrac{dy}{dt} = R_{\textsf{in}} - R_{\textsf{out}} \\ \\ \textsf{where:} \end{array} [/tex]

[tex] \begin{array}{l} \bullet \: R_{\textsf{in}} = \textsf{rate of the salt entering} \\ \quad \quad\: \: = \left({\footnotesize \begin{array}{c}\textsf{Concentration of} \\\textsf{salt inflow}\end{array}}\right) \times \small(\textsf{Input of brine}) \end{array} [/tex]

[tex] \begin{array}{l} \bullet \: R_{\textsf{out}} = \textsf{rate of the salt leaving} \\ \quad \quad\:\:\: = \left({\footnotesize \begin{array}{c}\textsf{Concentration of} \\\textsf{salt outflow}\end{array}}\right) \times \small(\textsf{Output of brine}) \end{array} [/tex]

[tex] \begin{array}{l} \textsf{Note that }\dfrac{y(t)}{1000} = x(t) \\ y(t) = 1000x(t) \implies \dfrac{dy}{dt} = 1000\dfrac{dx}{dt} \\ \\ R_{\textsf{in}} = 6(0.1) = 0.6\ \textsf{kg/min} \\ R_{\textsf{out}} = 6x(t)\ \textsf{kg/min} \\ \\ \textsf{Substituting these to the differential equation} \\ \textsf{above,} \\ \\ \implies 1000\dfrac{dx}{dt} = 0.6 - 6x \\ \\ \textsf{By separation of variables,} \\ \\ \displaystyle \int_0^{0.05} \dfrac{1000}{0.6 - 6x} dx = \int_0^T dt \\ \\ -\dfrac{1000}{6}\ln (0.6 - 6x) \Big |_0^{0.05} = T\\ \\ \therefore \boxed{T = 115.52\ \textsf{ mins}}, \textsf{where }T\textsf{ is the time taken} \\ \textsf{for the tank to reach a salt concentration of} \\ 0.05\textsf{ kg/L}.\end{array} [/tex]

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