Question

A tank contains 15,000 L of brine with 24 kg of dissolved salt. Pure water enters the tank at a rate of 150 L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate.How much salt is in the tank after t minutes

Answer 1

Explanation:

Let y(t) be the amount of salt in the tank after time t.

(A) Incoming rate = 0 (due to Pure water having no salt)

(B) Mixed solution comes out at 150 L/min. Initially the tank has 15,000 L of brine with 24 kg of salt.

concentration of salt at time t = y(t) / 15000 kg/L

Outgoing rate = y(t)/15000 * 150 = y(t) / 100

(C) we know that,

`(dy)/(dx) =(incoming\ rate) - (outgoing\ rate)`

`(dy)/(dx) =0-(y(t))/(100) = (-y(t))/(100)`

Separate variable and integrate

`\int {(dy)/(y) } = - \int {(1)/(100) } \, dt`

`ln|y|=-(1)/(100)t + D`

`y=e^(D) e^{(-t)/(100) }`

`y= Ce^{(-t)/(100) }\ [C=e^(D) ]`

At t= 0 , y(0) = 24 kg

`24=C\ e^(0)`

C= 24

(D) Therefore, the amount of salt in the tank after time t :

`y(t)=24e^{(-t)/(100) }\ kg`