Question

A tank contains 200litres of fluid in which 30grams of salt is dissolved. Brine containing 1gram of salt per litre is then pumped into the tank at a rate of;$\mathbf{4}\mathbf{L}\mathbf{/}\mathbf{min}$ the well-mixed solution is pumped out at the same rate. Find the number$\mathit{A}\mathbf{\left(}\mathit{t}\mathbf{\right)}$of grams of salt in the tank at time$\mathit{t}$.A

Short answer

The number of grams of salt in the tank at time is $A=\; 200\; -\; 170e^{-\frac{1}{\mathrm{31}}t}$.

Step-by-step solution

Define mixtures.

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

$\frac{\mathbf{d}\mathbf{A}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{(}\mathbf{}\mathbf{input}\mathbf{}\mathbf{rate}\mathbf{}\mathbf{of}\mathbf{}\mathbf{salt}\mathbf{}\mathbf{)}\mathbf{-}\mathbf{}\mathbf{(}\mathbf{output}\mathbf{}\mathbf{rate}\mathbf{}\mathbf{of}\mathbf{}\mathbf{salt}\mathbf{}\mathbf{)}\mathbf{=}{\mathit{R}}_{\mathit{i}\mathit{n}\mathbf{}}\mathbf{-}{\mathit{R}}_{\mathit{o}\mathit{u}\mathit{t}\mathbf{}}$

Solve for first order number of grams.

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

$\begin{array}{rcl}\frac{dA}{dt}& =& \mathrm{Input}\mathrm{rate}\mathrm{of}\mathrm{salt}-\mathrm{Input}\mathrm{rate}\mathrm{of}\mathrm{salt}\\ & & =F_{in}{C}_{in}-F_{out}{C}_{out}\end{array}$… (1)

Substitute the values$F_{in}=F_{out}=4L/\min ,C_{in}=1\mathrm{gram}/\mathrm{L},$and$C_{out}=\frac{\mathrm{Amount}\mathrm{of}\mathrm{salt}}{\mathrm{Volume}\mathrm{of}\mathrm{tank}}=\frac{A}{200}\mathrm{gram}/\mathrm{L}$in the equation (1).

$$\begin{gathered} \frac{dA}{dt}=4\mathrm{L}/\min \times 1\mathrm{gram}/\mathrm{L}-4\text{L/min}\times \frac{A}{200}\mathrm{gram}/\mathrm{L} \\ \frac{dA}{dt}=4-\frac{A}{50} \end{gathered}$$… (2)

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

$\begin{array}{rcl}& & \frac{dA}{4-\frac{A}{\mathrm{50}}}=dt\\ & & \int \frac{1}{\frac{200-A}{\mathrm{50}}}dA=\int dt\\ & & \int \frac{\mathrm{50}}{200-A}dA=\int dt\\ & & -\; 50\int \frac{-\; 1}{200-A}dA=\int dt\\ & & -\; 50ln(200\; -A)\; =t+c_1\\ & & ln(200\; -A)\; =\; -\frac{1}{\mathrm{50}}t+c_2\\ & & 200-A=e^{c_2}{e}^{-\frac{1}{\mathrm{50}}t}\end{array}$

Then, the equation becomes,

$\begin{array}{rcl}& & A=\; 200\; -e^{c_2}{e}^{-\frac{1}{\mathrm{30}}t}\\ & & =\; 200\; -{\mathrm{ce}}^{-\frac{1}{\mathrm{30}}t}\end{array}$… (3)

Obtain the values of constants.

To find the values of constants, apply the point$\text{(A,t) = (30 grams,0 minutes)}$in the equation (3), then

role="math" localid="1663923941503" $\begin{array}{rcl}& & 30\text{grams}=\; 200\; -c{e}^0\\ & & c=\; 200\; -\; 30\\ & & c=\; 170\end{array}$

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

$A=\; 200\; -\; 170e^{-\frac{1}{\mathrm{31}}t}$