A 500-gallon tank initially contains 220 gallons of pure distilled water. Brine containing 5 pounds of salt per gallon flows into the tank at the rate of 5 gallons per minute, and the well stirred mixture flows out of the tank at the rate of 5 gallons per minute. Find the amount of salt in the tank after 8 minutes.

Respuesta :

Answer: The amount of salt in the tank after 8 minutes is 36.52 pounds.

Step-by-step explanation:

Salt in the tank is modelled by the Principle of Mass Conservation, which states:

(Salt mass rate per unit time to the tank) - (Salt mass per unit time from the tank) = (Salt accumulation rate of the tank)

Flow is measured as the product of salt concentration and flow. A well stirred mixture means that salt concentrations within tank and in the output mass flow are the same. Inflow salt concentration remains constant. Hence:

[tex]c_{0} \cdot f_{in} - c(t) \cdot f_{out} = \frac{d(V_{tank}(t) \cdot c(t))}{dt}[/tex]

By expanding the previous equation:

[tex]c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt} + \frac{dV_{tank}(t)}{dt} \cdot c(t)[/tex]

The tank capacity and capacity rate of change given in gallons and gallons per minute are, respectivelly:

[tex]V_{tank} = 220\\\frac{dV_{tank}(t)}{dt} = 0[/tex]

Since there is no accumulation within the tank, expression is simplified to this:

[tex]c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt}[/tex]

By rearranging the expression, it is noticed the presence of a First-Order Non-Homogeneous Linear Ordinary Differential Equation:

[tex]V_{tank} \cdot \frac{dc(t)}{dt} + f_{out} \cdot c(t) = c_0 \cdot f_{in}[/tex], where [tex]c(0) = 0 \frac{pounds}{gallon}[/tex].

[tex]\frac{dc(t)}{dt} + \frac{f_{out}}{V_{tank}} \cdot c(t) = \frac{c_0}{V_{tank}} \cdot f_{in}[/tex]

The solution of this equation is:

[tex]c(t) = \frac{c_{0}}{f_{out}} \cdot ({1-e^{-\frac{f_{out}}{V_{tank}}\cdot t }})[/tex]

The salt concentration after 8 minutes is:

[tex]c(8) = 0.166 \frac{pounds}{gallon}[/tex]

The instantaneous amount of salt in the tank is:

[tex]m_{salt} = (0.166 \frac{pounds}{gallon}) \cdot (220 gallons)\\m_{salt} = 36.52 pounds[/tex]