Question

A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of 5 L/min. Brine that contains 0.04 kg of salt per liter of water enters the tank at a rate of 10 L/min. The solution is kept thoroughly mixed and drains from the tank at a rate of 15 L/min. How much salt is in the tank (a) after minutes and (b) after one hour?

Answer 1

(a) after t minutes

If $y(t)$ is the amount of salt (in kg) after $t$ minutes, then $y(0) = 0 \ (pure\ water)$ and the total amount of liquid in the tank remains constant at 1000 L.

$\displaystyle \frac{dy}{dt} \\ = \left( 0.05 \frac{\text{kg}}{\text{L}} \right) \left( 5 \frac{\text{L}}{\text{min}} \right) + \left( 0.04 \frac{\text{kg}}{\text{L}} \right)\left( 10 \frac{\text{L}}{\text{min}} \right) - \left( \frac{y(t)}{1000} \frac{\text{kg}}{\text{L}} \right)\left( 15\frac{\text{L}}{\text{min}} \right) \\ \\ = 0.25 + 0.40 - 0.015y = 0.65 - 0.015 y = \frac{130 - 3y}{200} \frac{\text{kg}}{\text{min}}$

$\text{Hence, } \displaystyle \int \frac{dy}{130 - 3y} = \int\frac{dt}{200}$

and $-\frac{1}{3}\ln|130 - 3y| = \frac{1}{200}t + C.$

Because $y(0) = 0$, we have $-\frac{1}{3} \ln 130 = C$

so

$-\frac{1}{3}\ln|130 - 3y| = \frac{1}{200}t -\frac{1}{3} \ln 130\ \Rightarrow \\ \\\ \ln|130 - 3y| = -\frac{3}{200}t + \ln 130 \ \Rightarrow \\ \\ |130 - 3y| = e^{-3t/200 + \ln 130} \ \Rightarrow \\ \\ |130 - 3y| = 130e^{-3t/200}$

Since $y$ is continuous, $y(0) = 0$, and the right-hand side is never zero, we deduce that $130 - 3y$ is always positive.

Thus, $130 - y = 130e^{-3t/200}$ and
$y = \frac{130}{3}\left( 1 - e^{-3t/200}\right) \text{ kg}$

(b)

After one hour,

$y = \frac{130}{3}\left( 1 - e^{-3 \cdot 60/200}\right) = \frac{130}{3}(1 - e^{-0.9} ) \approx 25.7 \text{ kg}$