Question

ZILLDIFFEQMODAP11 3.1.021. My Notes Ask Your Teacher A tank contains 150 liters of fluid in which 10 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

Answer:

Therefore, the number of grams of salt in the tank at time t is $A(t) = 150-140 e^{-\frac{t}{30} }$

Explanation:

Given:

Tank A contain $V_{1} = 150$ lit

Rate $\alpha = 5 \frac{L}{min}$

Dissolved salt $A = 10$ gm

Salt pumped in one minute is $4 \frac{L}{min}$

Salt pumped out is $\frac{5L}{150L} = \frac{1}{30}$ of initial amount added salt.

To find $A(t)$

  $\frac{dA}{dt} = Rate _{in} - Rate _{out}$

  $A' = 5 - \frac{A}{30}$

  $A' + \frac{A}{30} = 5$

Solving above equation,

  $I .F = e^{\int\limits {p} \, dt }$

   $y = e^{\int\limits {\frac{1}{30} } \, dt }$

   $y = e^{\frac{t}{30} }$

$(Ae^{\frac{t}{30} } )' = 5 e^{\frac{t}{30} } + c$

Integrating on both side,

$Ae^{\frac{t}{30} } = 5 \times 30 e^{\frac{t}{30} } +c$

Add $e^{-\frac{t}{30} }$ on above equation,

 $A = 150 + ce^{-\frac{t}{30} }$

Here given in question,$A(t=0) = 10$

 $10 =150 +c$

   $c = -140$

Put value of constant in above equation, and find the number of grams of salt in the tank at time t.

 $A(t) = 150-140 e^{-\frac{t}{30} }$

Therefore, the number of grams of salt in the tank at time t is $A(t) = 150-140 e^{-\frac{t}{30} }$