Answer:
[tex]c(t) = 0.044\cdot e ^{-0.066\cdot t}[/tex]
Step-by-step explanation:
The quantity of salt inside the tank is modelled after the Principle of Mass Conservation:
Salt
[tex]\dot m_{in, salt} - \dot m_{out, salt} = \frac{dm_{tank,salt}}{dt}[/tex]
Water
[tex]\dot m_{in,water} - \dot m_{out,water} = \frac{dm_{tank,water}}{dt}[/tex]
Given that water is an incompressible fluid, the expression can be simplified into the following expression:
[tex]\dot V_{in, water} - \dot V_{out,water} = \frac{dV_{tank, water}}{dt}[/tex]
Both flows have the same rate and tank can be modelled as a steady state system.
[tex]\dot V_{in, water} - \dot V_{out,water} = 0[/tex]
The expression for salt concentration in the tank is:
-[tex]-\dot V_{tank}\cdot c = V_{tank} \cdot \frac{dc}{dt}[/tex]
After some handling, the following homogeneous first-order linear differential equation is found:
[tex]\frac{V_{tank}}{\dot V_{tank}} \cdot \frac{dc}{dt} + c = 0[/tex]
Where [tex]c (0) = 0.044\,\frac{lbm}{gal}[/tex]. The solution is obtained by using Laplace transforms:
[tex]\frac{V_{tank}}{\dot V_{tank}} \cdot \left[s\cdot C(s) - c(0)\right] + C(s) = 0[/tex]
[tex]\left(\frac{V_{tank}}{\dot V_{tank}}\cdot s + 1\right)\cdot C(s) = \frac{V_{tank}}{\dot V_{tank}}\cdot c(0)[/tex]
[tex]C(s) = \frac{\frac{V_{tank}}{\dot V_{tank}}\cdot c(0) }{\left(\frac{V_{tank}}{\dot V_{tank}} \right)\cdot \left(s + \frac{\dot V_{tank}}{V_{tank}} \right)}[/tex]
[tex]c(t) = c(0) \cdot e^{-\frac{\dot V_{tank}}{V_{tank}}\cdot t }[/tex]
Where [tex]\frac{\dot V_{tank}}{V_{tank}} = 0.066\,min^{-1}[/tex].
The formula for the concentration of salt in the tank is:
[tex]c(t) = 0.044\cdot e ^{-0.066\cdot t}[/tex]
