Question

A tank contains 240 liters of fluid in which 10 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 6 L/min; the well-mixed solution is pumped out at the same rate. Find the number A(t) of grams of salt in the tank at time t.

Answer 1

Answer:

$\boxed{240 - 230e^{-\frac{t}{40}}}$

Explanation:

$\text{Let A = mass of salt after t min}\\\text{and }r_{i} = \text{rate of salt coming into talk}\\\text{and }r_{o} =\text{rate of salt going out of tank}$

1. Set up an expression for the rate of change of salt concentration.

$\dfrac{\text{d}A}{\text{d}t} = r_{i} - r_{o}\\\\r_{i} = \dfrac{\text{6 L}}{\text{1 min}} \times \dfrac{\text{1 g}}{\text{1 L}} = \text{6 g/min}\\\\r_{o} = \dfrac{\text{6 L}}{\text{1 min}} \times \dfrac {A\text{ g}}{\text{240 L}} =\dfrac{x}{40}\text{ g/min}\\\\\dfrac{\text{d}A}{\text{d}t} = 6 - \dfrac{x}{40}$

2. Integrate the expression

$\dfrac{\text{d}A}{\text{d}t} = \dfrac{240 - x}{40}\\\\\dfrac{\text{d}A}{240 - A} = \dfrac{\text{d}t}{40}\\\\\int \frac{\text{d}A}{240 - A} = \int \frac{\text{d}t}{40}\\\\-\ln |240 - A| = \frac{t}{40} + C$

3. Find the constant of integration

$-\ln |240 - A| = \frac{t}{40} + C\\\\\text{At t = 0, A = 10, so}\\\\-\ln |240 - 10| = \frac{0}{40} + C\\\\C = -\ln 230$

4. Solve for A as a function of time.

$\text{The integrated rate expression is}-\ln |240 - A| = \frac{t}{40} - \ln 230\\\\\text{Solve for } A\\\\\ln|240 - A| = \ln 230 - \frac{t}{40}\\\\|240 - A| = 230e^{-\frac{t}{40}}\\\\240 - A = \pm 230e^{-\frac{t}{40}}\\\\x = 240 \pm 230e^{-\frac{t}{40}}\\\\A(0) = 10 \text{ so we choose the negative sign}\\\\x = \boxed{\mathbf{240 - 230e^{-\frac{t}{40}}}}$

The diagram shows A as a function of time. The mass of salt in the tank starts at 10 g and increases asymptotically to 240 g.