At time t 0 a 300L tank contains 300 Liters of water with 5

At time t = 0, a 300-L tank contains 300 Liters of water with 50 kg of salt dissolved in it. A brine solution with concentration of 3 kg/L is pumped into the tank at the rate of 4 L/min, and the well-mixed solution is pumped out of the tank at the rate of 2 L/min. Find an expression for the amount of salt in the tank at time t. How much salt is there in the tank, when it is full?

Solution

Let us define the function s(t) = kgs of salt in the tank at time t minutes.

We don’t know what s(t) is, but we do know its derivative: s’(t) = the rate at which the amount of salt in the tank is changing = (rate of salt going in) (rate of salt going out)

The quantity of salt is going into the tank per minute = (3kgs of salt/ liter) × 4 liters/ minute = 12 kgs of salt/ minute.

The quantity of salt per liter in the tank (which is kept thoroughly mixed) is s(t) kgs/ 300 liters. Further, the solution is leaving the tank at the rate of 2 liters/ minute. Therefore, the rate at which salt is leaving the tank is s(t) kgs of salt /300 liters × 2 liters/ minute =( 2/300) s(t) kgs of salt/ minute.

Therefore, the differential equation describing this situation is s’(t) = (rate of salt going in) (rate of salt going out) = 12 – (2/300 )s(t) = 12 – 0.0067 s(t) i.e. ds/dt = 12 – 0.0067 s(t) or, ds/dt + 0.0067 s(t) = 12.

We know that the solution of the equation y’ + ay = g(x) is y = e^-ax e^ax g(x) dx + De^-ax. Therefore, for this problem, s(t) = e^-0.0067t *12e^0.015t + De^0.015t i.e. s(t) = 12 + De^0.015t . Now, using the initial condition s(0) = 50, we have 50 = 12 + D so that D = 50 -12 = 38. Then s(t) = 12 + 38e^0.015t . Further, initially, when t = 0, the 300 liter tank contains 300 liters of water with 50 kg salt dissolved in it. Therefore, there is 50 kg of salt in the tank when it is full.