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
