Question

A tank contains 100 kg of salt and 1000 L of water. A solution of a concentration 0.05 kg of salt per liter enters a tank at the rate 99 L/min. The solution is mixed and drains from the tank at the same rate.
Required:
a. Find the amount of salt in the tank after 3.5 hours.
b. Find the concentration of salt in the solution in the tank as time approaches infinity.

Answer 1

Correct question is;

A tank contains 100 kg of salt and 1000 L of water. A solution of a concentration 0.05 kg of salt per liter enters a tank at the rate 9 L/min. The solution is mixed and drains from the tank at the same rate.

Required:

a. Find the amount of salt in the tank after 3.5 hours.

b. Find the concentration of salt in the solution in the tank as time approaches infinity.

Answer:

A) y(3.5) = 40.11 kg of salt

B) Concentration as time approaches infinity = 0.05 kg/l

Step-by-step explanation:

We are given;

Mass of salt; m_s = 50 kg

Volume of water; v_w = 1000 L

Rate at which salt enters = 0.05 kg/L × 9 L/min. = 0.45 kg/min

Since the solution drains at same rate as it enters, then;

Rate at which salt goes out = 9y/1000

Where y is the concentration in the tank.

Thus, the differential equation of the amount of water in the tank will be;

dy/dt = Rate at which salt enters - Rate at which salt goes out

dy/dt = 0.45 - (9y/1000)

Simplifying this gives;

dy/dt = (9/1000)(50 - y)

Rearranging, we have;

dy/(50 - y) = dt(9/1000)

Integrating both sides gives;

In(50 - y) = 9t/1000 + A

If we do exponents of both sides, we will get;

50 - y = Ae^(-9t/1000)

At initial conditions, y = 0 and t = 0.

Thus;

A = 50

Thus, quantity of salt in tank will be written as;

50 - y = 50e^(-9t/1000)

Making y the subject gives;

y = 50 - 50e^(-9t/1000)

At t = 3.5 hours = 210 minutes

y = 50 - 50e^(-9 × 180/1000)

y(3.5) = 40.11 kg of salt

As t approaches infinity, it means t will be zero. Thus;

y = 50 - 50e^(-9 × 0/1000)

y = 50 kg

Concentration = 50/1000 = 0.05 kg/l