Question

A tank contains 150 liters of fluid in which 40 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 3 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:

The number A(t) of grams of salt in the tank at time t is $A(t) = 150 - 110 e^{-\frac{t}{50} }$

Explanation:

Knowing

$\frac{dA}{dt} = Rin - Rout$

First we have to find the Rin and Rout

Rin = (concentration of the salt inflow) * (input rate of brine)

Rin = 1 g/L * 3 L/min = 3 g/L

Rout = (concentration of the salt outflow) * (output rate of brine)

Rout = ($\frac{A(t)}{150} g/L$) * (3 L/min) = $\frac{A(t)}{50} g/min$

Substituting this results

$\frac{dA}{dt}$ = 3 - $\frac{A(t)}{50}$ --> $\frac{dA}{dt} + \frac{1}{50} A(t) = 4$

Thus, integration factors is

$e^{ \int\limits^._. {\frac{1}{50} } \, dt } = e^{\frac{t}{50} }$

$e^{\frac{t}{50} } \frac{dA}{dt} + \frac{1}{50} e^{\frac{t}{50}$ A(t) = 4 $e^{\frac{t}{50} }$

$e^{\frac{t}{50} } A(t) = \int\limits^._. {3 e^{\frac{t}{50} } } \, dt\\ \\A(t) = 150 + c e^{-\frac{t}{50} }$

Applying the initial conditions

A(0) = 40

c = 150 - 40 = 110

Now, substitute this result in the solution to get

$A(t) = 150 - 110 e^{-\frac{t}{50} }$