Question

In Problem 23, what is the concentration $\mathbf{c}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$of the salt in the tank at time t? At t=5min? What is the concentration of the salt in the tank after a long time, that is, as$\mathbf{t}\mathbf{\to }\mathbf{\infty }$? At what time is the concentration of the salt in the tank equal to one-half this limiting value?

Short answer

The concentration of the salt and the required time are,

$$\begin{gathered} \mathrm{C}=2-2{\mathrm{e}}^{-\frac{1}{1+0^2}\mathrm{t}}\mathrm{Ib}/\mathrm{gallons} \\ \mathrm{C}=0.0975\mathrm{Ib}/\mathrm{gallon} \\ \mathrm{C}=2\mathrm{Ib}/\mathrm{gallon} \\ \mathrm{t}=69.3\mathrm{minutes} \end{gathered}$$

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 concentration of salt at a time t.

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

As the volume of tank is 500 galloons, then the concentration of salt at a time t is,

$\begin{array}{rcl}\mathrm{C}& =& \frac{\mathrm{Number}\mathrm{of}\mathrm{bounds}\mathrm{of}\mathrm{salt}}{\mathrm{Volume}\mathrm{of}\mathrm{tank}}\\ & =& \frac{\mathrm{A}\left(\mathrm{t}\right)}{\mathrm{V}}\\ & =& \frac{1000-1000{\mathrm{e}}^{-\frac{1}{10\mathrm{i}}\mathrm{t}}}{500}\mathrm{Ib}/\mathrm{gallon}\\ & & \end{array}$

$\mathrm{C}=2-2{\mathrm{e}}^{-\frac{1}{100}\mathrm{t}}\mathrm{Ib}/\mathrm{gallon}$… (1)

Solve for concentration of salt at a time t = 5.

As t = 5, then the concentration of salt at a time using (1) is,

$\begin{array}{rcl}\mathrm{C}& =& 2-2{\mathrm{e}}^{-\frac{1}{100}\times 5}\mathrm{Ib}/\mathrm{gallon}\\ & =& 2-2{\mathrm{e}}^{-\frac{1}{21}}\mathrm{Ib}/\mathrm{gallon}\\ & =& (2-2\times 0.95123)\mathrm{Ib}/\mathrm{gallon}\\ & =& (2-1.9025)\mathrm{Ib}/\mathrm{gallon}\\ & =& 0.0975\mathrm{Ib}/\mathrm{gallon}\end{array}$

Solve for concentration of salt at a time t=∞.

As$\mathrm{t}=\infty$, then the concentration of salt at a time using (1) is,

$\begin{array}{rcl}\mathrm{C}& =& 2-2{\mathrm{e}}^{-\frac{1}{\mathrm{IIn}}\times \infty }\mathrm{Ib}/\mathrm{gallon}\\ & =& 2-2{\mathrm{e}}^{-\infty }\mathrm{Ib}/\mathrm{gallon}\\ & =& (2-0)\mathrm{Ib}/\mathrm{gallon}\\ & =& 2\mathrm{Ib}/\mathrm{gallon}\end{array}$

Solve for time for concentration is one half.

Substitute the value$C=1\mathrm{Ib}/\mathrm{galloon}$ in equation (1) to find the required time.

$\begin{array}{rcl}1\mathrm{Ib}/\mathrm{gallons}& =& 2-2{\mathrm{e}}^{-\frac{1}{100}\mathrm{t}}\mathrm{Ib}/\mathrm{gallons}\\ 2-1& =& 2{\mathrm{e}}^{-\frac{1}{100}\mathrm{t}}\\ {\mathrm{e}}^{-\frac{1}{100}\mathrm{t}}& =& \frac{1}{2}\\ {\mathrm{lne}}^{-\frac{1}{1000}\mathrm{t}}\mathrm{t}& =& \ln \left(\frac{1}{2}\right)\\ -\frac{1}{100}\mathrm{t}& =& \ln \left(\frac{1}{2}\right)\\ \mathrm{t}& =& -100\ln \left(\frac{1}{2}\right)\\ & =& -100\times -0.693\\ & =& 69.3\mathrm{minutes}\end{array}$