Question

A very large tank initially contains 100L of pure water. Starting at time t=0 a solution with a salt concentration of 0.5kg/L is added at a rate of 6L/min. The solution is kept thoroughly mixed and is drained from the tank at a rate of 4L/min.How much salt is in the tank after 30 minutes?
Answer (in kilograms)

Answer 1

Given:

A large tank initially contains 100L of pure water.

At time t = 0, a solution with a salt concentration of 0.5kg/L is added at a rate of 6L/min.

The solution is kept thoroughly mixed and is drained from the tank at a rate of 4L/min.

To Find:

How much salt is in the tank after 30 minutes.

Solution:

Let y(t) = amount of salt (in kg) in the tank after t minutes.

Then,

dy  / dt = (rate in) − (rate out)

= (0.5)(6) −  (y(t)  / 100) × 4

= 3 − (y /0.04)

= (0.12 − y) /0.04.

This is a separable differential equation.

So we  can write,

$\int\ \frac{dy}{0.12-y} = \int\ \frac{dt}{0.12}$  

By Anti differentiating, we get

− ln |0.12 − y| = t/0.12 + C.

Plug in the initial condition y(0) = 0 to get,

− ln 0.12 = C,

and do some  algebra to get the explicit solution

ln |0.12 − y| = ln 0.12 − t/0.12

|0.12 − y| = e ∧(ln 0.12) × e ∧(-t/0.12)  = 0.12e  ∧−t/0.12.

Since y < 0.12, we may drop the absolute value symbols:

y = 0.12 − 0.12e  ∧(−t/0.12) = 0.12(1 − e  ∧(−t/0.12))

After 30 minutes, the quantity of salt in the tank will be

y(30) = 0.12(1 − e  ∧(-30/0.12)) = 0.12 (1 − e  ∧−250 ) kg

Hence, the salt left in the tank after 30 minutes is 0.12 (1 − e  ∧−250 ) kg.