Question

A tank contains 9,000 L of brine with 18 kg of dissolved salt. Pure water enters the tank at a rate of 90 L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate.
(a) How much salt is in the tank after t minutes?
(b) How much salt is in the tank after 20 minutes?

Answer 1

Let x(t) denote the amount of salt (in kg) in the tank at time t. The tank starts with 18 kg of salt, so x (0) = 18.

The solution is drained from the tank at a rate of 90 L/min, so that the amount of salt in the tank changes according to the differential equation

dx(t)/dt = - (x(t) kg)/(9000 L) × (90 L/min) = -1/100 x(t) kg/min

or, more succintly,

x' = -1/100 x

This equation is separable as

dx/x = -1/100 dt

Integrating both sides gives

∫ dx/x = -1/100 ∫ dt

ln|x| = -1/100 t + C

x = exp(-1/100 t + C )

x = C exp(-t/100)

(a) Using the initial condition x (0) = 18, we find

18 = C exp(0)   ==>   C = 18

so that

x(t) = 18 exp(-t/100)

(b) After 20 minutes, we have

x (20) = 18 exp(-20/100) = 18 exp(-1/5) ≈ 14.74

so the tank contains approximately 14.74 kg of salt after this time.