Question

A large tank is filled to capacity with 500gallons of pure water. Brine containing 2pounds of salt per gallon is pumped into the tank at a rate of$\mathbf{5}\mathbf{}\mathbf{gal}\mathbf{/}\mathbf{min}$. The well-mixed solution is pumpedout at the same rate. Find the number$\mathit{A}\mathbf{\left(}\mathit{t}\mathbf{\right)}$of pounds of salt inthe tank at time$\mathit{t}$.

Short answer

The number of pounds of salt in the tank at time is ${\text{A(t) = 1000 - 1000e}}^{\text{-}\frac{\text{1}}{100}\text{t}}$.

Step-by-step solution

Define mixtures.

The mixing of two fluids sometimes gives rise to a linear first-order differential equation.

$\frac{dA}{dt}=\left(\text{input rate of salt}\right)-\text{(output rate of salt})\mathbf{=}{\mathit{R}}_{\mathbf{\text{in}}}\mathbf{-}{\mathit{R}}_{\mathbf{\text{out}}}$

Solve for first order number of pounds.

Let the linear equation with the amount of salt in pounds as A be,

$$\begin{gathered} \frac{\text{dA}}{\text{dt}}\text{= Input rate of salt - Input rate of salt} \\ {\text{=F}}_{\text{in}}{\text{C}}_{\text{in}}{\text{-F}}_{\text{out}}{\text{C}}_{\text{out}} \end{gathered}$$… (1)

Substitute the values${\mathrm{F}}_{\mathrm{in}}={\mathrm{F}}_{\mathrm{out}}=5\mathrm{gallon}/\min ,{\mathrm{C}}_{\mathrm{in}}=2\mathrm{Ib}/\mathrm{gallon},$ and${\text{C}}_{\text{out}}\text{=}\frac{\text{Amount of salt}}{\text{Volume of tank}}\text{=}\frac{\text{A}}{\text{500}}\text{Ib/gallon}$ in the equation (1).

$$\begin{gathered} \frac{\mathrm{dA}}{\mathrm{dt}}=5\mathrm{gallon}/\min \times 2\mathrm{Ib}/\mathrm{gallon}-5\mathrm{gallon}/\min \times \frac{\mathrm{A}}{500}\mathrm{Ib}/\mathrm{gallon} \\ \frac{\mathrm{dA}}{\mathrm{dt}}=10-\frac{\mathrm{A}}{100} \end{gathered}$$… (2)

As the equation (2) is linear and separable, so integrate the equation and separate the variables.

$$\begin{gathered} \frac{\text{dA}}{\text{10 -}\frac{\text{A}}{100}}\text{= dt} \\ \int \frac{1}{\frac{1000-A}{100}}\text{dA =}\int \text{dt} \\ \int \frac{100}{1000-A}\text{dA =}\int \text{dt} \\ \text{- 100}\int \frac{\text{- 1}}{\text{1000 - A}}\text{dA =}\int \text{dt} \\ {\text{- 100ln(1000 - A) = t +c}}_{\text{1}} \\ \text{ln(1000 - A) = -}\frac{\text{1}}{100}{\text{t +c}}_{\text{2}} \\ {\text{1000 - A =e}}^{{\text{c}}_{\text{2}}}{e}^{\text{-}\frac{\text{1}}{100}\text{t}} \end{gathered}$$

Then, the equation becomes,

$$\begin{gathered} {\text{A(t) = 1000 -e}}^{{\text{c}}_{\text{2}}}{\text{e}}^{\text{-}\frac{\text{1}}{100}\text{t}} \\ {\text{= 1000 - ce}}^{\text{-}\frac{\text{1}}{\text{100}}\text{t}} \end{gathered}$$… (3)

Obtain the values of constants.

To find the values of constants, apply the point$\text{(A,t) = (0 Ib,0 minutes)}$ in the equation (3), then

$$\begin{gathered} {\text{0 Ib = 1000 - ce}}^{\text{0}} \\ \text{c = 1000} \end{gathered}$$

Substitute the value of$c$ in the equation (3).

${\text{A(t) = 1000 - 1000e}}^{\text{-}\frac{\text{1}}{100}\text{t}}$