Answer:
The quantity of salt at time t is [tex]m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })[/tex], where t is measured in minutes.
Step-by-step explanation:
The law of mass conservation for control volume indicates that:
[tex]\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}[/tex]
Where mass flow is the product of salt concentration and water volume flow.
The model of the tank according to the statement is:
[tex](0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}[/tex]
Where:
[tex]c[/tex] - The salt concentration in the tank, as well at the exit of the tank, measured in [tex]\frac{pd}{gal}[/tex].
[tex]\frac{dc}{dt}[/tex] - Concentration rate of change in the tank, measured in [tex]\frac{pd}{min}[/tex].
[tex]V[/tex] - Volume of the tank, measured in gallons.
The following first-order linear non-homogeneous differential equation is found:
[tex]V \cdot \frac{dc}{dt} + 6\cdot c = 3[/tex]
[tex]60\cdot \frac{dc}{dt} + 6\cdot c = 3[/tex]
[tex]\frac{dc}{dt} + \frac{1}{10}\cdot c = 3[/tex]
This equation is solved as follows:
[tex]e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }[/tex]
[tex]\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }[/tex]
[tex]e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt[/tex]
[tex]e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C[/tex]
[tex]c = 30 + C\cdot e^{-\frac{t}{10} }[/tex]
The initial concentration in the tank is:
[tex]c_{o} = \frac{10\,pd}{60\,gal}[/tex]
[tex]c_{o} = 0.167\,\frac{pd}{gal}[/tex]
Now, the integration constant is:
[tex]0.167 = 30 + C[/tex]
[tex]C = -29.833[/tex]
The solution of the differential equation is:
[tex]c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }[/tex]
Now, the quantity of salt at time t is:
[tex]m_{salt} = V_{tank}\cdot c(t)[/tex]
[tex]m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })[/tex]
Where t is measured in minutes.
