Question

A tank contains 200 liters of fluid in which 50 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 5 L/min; the well-mixed solution is pumped out at the same rate. Find the number A(t) of grams of salt in the tank at time t.

Answer 1

At the start, the tank contains A(0) = 50 g of salt.

Salt flows in at a rate of

(1 g/L) * (5 L/min) = 5 g/min

and flows out at a rate of

(A(t)/200 g/L) * (5 L/min) = A(t)/40 g/min

so that the amount of salt in the tank at time t changes according to

A'(t) = 5 - A(t)/40

Solve the ODE for A(t):

A'(t) + A(t)/40 = 5

e^(t/40) A'(t) + e^(t/40)/40 A(t) = 5e^(t/40)

(e^(t/40) A(t))' = 5e^(t/40)

e^(t/40) A(t) = 200e^(t/40) + C

A(t) = 200 + Ce^(-t/40)

Given that A(0) = 50, we find

50 = 200 + C  ==>  C = -150

so that the amount of salt in the tank at time t is

A(t) = 200 - 150 e^(-t/40)