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,
[tex]\frac{dy}{dx} =(incoming\ rate) - (outgoing\ rate)[/tex]
[tex]\frac{dy}{dx} =0-\frac{y(t)}{100} = \frac{-y(t)}{100}[/tex]
Separate variable and integrate
[tex]\int {\frac{dy}{y} } = - \int {\frac{1}{100} } \, dt[/tex]
[tex]ln|y|=-\frac{1}{100}t + D[/tex]
[tex]y=e^{D} e^{\frac{-t}{100} }[/tex]
[tex]y= Ce^{\frac{-t}{100} }\ [C=e^{D} ][/tex]
At t= 0 , y(0) = 24 kg
[tex]24=C\ e^{0}[/tex]
C= 24
(D) Therefore, the amount of salt in the tank after time t :
[tex]y(t)=24e^{\frac{-t}{100} }\ kg[/tex]
