Question

Solve Problem 23 under the assumption that the solution is pumped out at a faster rate of $\mathbf{10}\mathbf{}\mathbf{gal}\mathbf{/}\mathbf{min}$. When is the tank empty?

Short answer

The number of pounds of salt in the tank at time is ${\text{A(t) = 1000 - 1000e}}^{\text{-}\frac{\text{1}}{100}\text{t}}$. The time at which the large tank is empty from the function is role="math" localid="1663927020053" $\mathrm{t}=100\mathrm{minutes}$.

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${\text{F}}_{\text{in}}{\text{= 5 gallon/min,F}}_{\text{out}}{\text{= 10 gallon/min,C}}_{\text{in}}\text{= 2 Ib/gallon,}$ and${\mathrm{C}}_{\mathrm{out}}=\frac{\mathrm{Amountofsalt}}{\mathrm{Volumeoftank}}=\frac{\mathrm{A}\left(\mathrm{t}\right)}{\mathrm{V}-({\mathrm{F}}_{\mathrm{out}}-{\mathrm{F}}_{\mathrm{in}})\mathrm{t}}\mathrm{Ib}/\mathrm{gallon}=\frac{\mathrm{A}}{500-5\mathrm{t}}\mathrm{Ib}/\mathrm{gallon}$ in the equation (1).

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

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

role="math" localid="1663927181542" $\begin{array}{rcl}\mathrm{F}& =& {\mathrm{e}}^{\int \mathrm{p}\left(\mathrm{t}\right)\mathrm{dt}}\\ & =& {\mathrm{e}}^{\int \frac{2}{100-\mathrm{t}}\mathrm{dt}}\\ & =& {\mathrm{e}}^{-2\int \frac{-1}{100-\mathrm{t}}\mathrm{dt}}\\ & =& {\mathrm{e}}^{-2\ln (100-\mathrm{t})}\\ & =& (100-\mathrm{t}{)}^{-2}\\ & =& \frac{1}{{(100-\mathrm{t})}^2}\end{array}$

Multiply the equation by the IF.

Let,

$\begin{array}{rcl}\frac{1}{{(100-\mathrm{t})}^2}\frac{\mathrm{dA}}{\mathrm{dt}}+\left(\frac{1}{{(100-\mathrm{t})}^2}\right)\frac{2}{100-\mathrm{t}}\mathrm{A}& =& 10\left(\frac{1}{{(100-\mathrm{t})}^2}\right)\\ \frac{1}{{(100-\mathrm{t})}^2}\frac{\mathrm{dA}}{\mathrm{dt}}+\frac{2}{{(100-\mathrm{t})}^3}\mathrm{A}& =& \frac{10}{{(100-\mathrm{t})}^2}\\ \frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{1}{{(100-\mathrm{t})}^2}\mathrm{A}\right)& =& \frac{10}{{(100-\mathrm{t})}^2}\end{array}$

Solve by integration,

$\begin{array}{rcl}\int \mathrm{d}\left(\frac{1}{{(100-\mathrm{t})}^2}\mathrm{A}\right)& =& 10\int {(100-\mathrm{t})}^{-2}\mathrm{dt}\\ \frac{1}{{(100-\mathrm{t})}^2}\mathrm{A}& =& \frac{10}{(100-\mathrm{t})}+\mathrm{C}\\ \mathrm{A}\left(\mathrm{t}\right)& =& (100-\mathrm{t}{)}^2\frac{10}{(100-\mathrm{t})}+\mathrm{C}(100-\mathrm{t}{)}^2\\ & =& \frac{10{(100-\mathrm{t})}^2}{(100-\mathrm{t})}+\mathrm{C}(100-\mathrm{t}{)}^2\end{array}$

$\mathrm{A}\left(\mathrm{t}\right)=10(100-\mathrm{t})+\mathrm{C}(100-\mathrm{t}{)}^2$… (3)

Obtain the values of constants.

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

$\begin{array}{rcl}Ib& =& 10(100-0)+C(100-0{)}^2\\ 1000+10,000C& =& 0\\ c& =& -\frac{1000}{10,000}\\ & =& -\frac{1}{10}\end{array}$

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

$\mathrm{A}\left(\mathrm{t}\right)=10(100-\mathrm{t})-\frac{1}{10}(100-\mathrm{t}{)}^2$

Obtain the time at which the large tank is empty.

To obtain the time at which the large tank is empty from the function is:

$\begin{array}{rcl}\mathrm{V}& =& 500-5\mathrm{t}\\ 5\mathrm{t}& =& 500-0\\ \mathrm{t}& =& \frac{500}{5}\\ & =& 100\mathrm{minutes}\end{array}$