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

Answer:

Step-by-step 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,

$\frac{dy}{dx} =(incoming\ rate) - (outgoing\ rate)$

$\frac{dy}{dx} =0-\frac{y(t)}{100} = \frac{-y(t)}{100}$

Separate variable and integrate

$\int {\frac{dy}{y} } = - \int {\frac{1}{100} } \, dt$

$ln|y|=-\frac{1}{100}t + D$

$y=e^{D} e^{\frac{-t}{100} }$

$y= Ce^{\frac{-t}{100} }\ [C=e^{D} ]$

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 :

$y(t)=24e^{\frac{-t}{100} }\ kg$