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

The amount of salt in the tank after 3.5 hours is -939.5 kg. The concentration of salt in the solution in the tank as time approaches infinity is 1 kg/L.

Step-by-step explanation:

To find the amount of salt in the tank after 3.5 hours, we need to calculate the total amount of salt that has entered and drained from the tank in that time period.

The rate at which the solution enters and drains from the tank is 99 L/min. Therefore, in 3.5 hours (which is equal to 210 minutes), a total of 99 L/min * 210 min = 20790 L of solution enters and drains from the tank.

The concentration of the solution is 0.05 kg/L. Therefore, the amount of salt that enters and drains from the tank in 20790 L of solution is 0.05 kg/L * 20790 L = 1039.5 kg. Subtracting this from the initial amount of salt in the tank (100 kg), we get the amount of salt in the tank after 3.5 hours as 100 kg - 1039.5 kg = -939.5 kg.

The concentration of salt in the solution in the tank as time approaches infinity can be calculated using the concept of the limiting concentration. As the solution enters and drains from the tank at the same rate, the amount of salt in the tank will eventually reach a constant value. This constant value represents the concentration of salt in the solution in the tank as time approaches infinity.

Since the total amount of salt entering and draining from the tank is 20790 kg (as calculated earlier), and the total volume of solution entering and draining from the tank is 20790 L, the concentration of salt in the solution in the tank as time approaches infinity is 20790 kg / 20790 L = 1 kg/L.

Answer 2

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